Introduction to Volume & Disk Method: Videos & Practice Problems

To find the volume of a three-dimensional solid, we can use integration by considering the area of cross sections. For solids of revolution, formed by rotating a function around an axis, the cross sections are circular, leading to the volume integral $V^2=\int {\pi }_r{r}^{2}dx$ or $\int {\pi }_r{r}^{2}dy$ depending on the axis of rotation. Understanding the radius function and setting appropriate bounds is crucial for accurate volume calculation.

Introduction to Cross Sections

Earlier in this course, we learned how to find the area underneath the curve of a function by setting up an integral. And in doing that, we were really taking our two-dimensional area and breaking it up into a bunch of small rectangles. We are then adding the area of all of those rectangles together to get that total area. But what if instead we were asked to find the volume of a three-dimensional solid? That's exactly what we're going to dive into in this video.

And just like working with our two-dimensional area where we cut this up into a bunch of small rectangles, we're going to take our three-dimensional solid and cut it up into a bunch of thin slices. Now let's not waste any time here and jump right into our example. We're asked to set up an integral for the volume of a solid that has a base formed by the function f(x) = 4 - x² on the interval from zero to two, and we're told that this solid has square cross sections. So let's start here by coming down to our graph and taking a look at our function. We have our function here 4 - x² on the interval from zero to two.

So this shaded area represents the base of our solid. So how does this then become three-dimensional? Well, if we take our two-dimensional xy plane that we know and love and tilt it backwards, then extending that area upwards, we have this three-dimensional solid. Then just like our area being cut up into a bunch of rectangles, we can cut our solid into a bunch of slices. If we then just focus on one slice out of our solid, we have this sort of prism shape.

If we then take this prism and face it towards us, we see this square cross section as was told to us in our problem. So now that we have a better idea of what this solid actually looks like, how can we then find its volume? When working with our two-dimensional area, we were able to cut one rectangle out of there and find its area by multiplying its height times its width. Now that we're working with a three-dimensional solid, we're going to take one slice out of our solid and focus on finding its volume. Now in order to find the volume of just a single slice out of my solid here, we want to take the area of that slice's cross section and multiply it by its width.

Now just as a reminder, a cross section is a two-dimensional shape that we get from cutting into a three-dimensional solid. So if I had something like a block of cheese and I were to slice up that block of cheese and slap one of those slices onto my sandwich, I would see that it's shaped like a square because it has square cross sections. Now, in this particular problem, we are also working with these square cross sections. So coming back down to our slice here, in order to find the volume of this single slice, we want to take our area function, which we're going to call here a(x), and multiply it by the width of our slice. Here, the width of our slice is dx because we're taking these infinitely thin slices along our x-axis.

So with that width in mind, we just need to figure out what that area function is. Because we're working with squares here, I have my front-facing cross section of my square with these particular side lengths. Now coming back over here to our three-dimensional shape, if I look at the base of this slice, I can see that it's bounded by my function f(x). So that's exactly what that bottom side length is. It's f(x).

So because this is a square, if that one side length is f(x), all of my side lengths are f(x). So this will now allow me to find my area function, a(x). So in this case, a(x) is going to be equal to f(x) times f(x), which is just f(x)². Now we know exactly what f(x) is. It's 4 - x², so that makes this area function (4 - x²)².

So now that we have our area function, we can go ahead and actually find the volume of this slice by taking our area function and multiplying it by that width. So taking (4 - x²)² and multiplying it by our width here of dx. This gives me the volume of a single slice, so just this one slice out of my solid. But remember, we want to find the volume of this whole entire solid. So we can do this by taking what we have here and integrating it on our interval from zero to two to account for all of those thin slices.

So this integral ultimately gives us the volume of this three-dimensional solid that we have on our graph here. So when we were working with finding the area underneath the curve of a function, we were able to integrate on our interval from a to b our function f(x) dx. Now when working with finding the volume of a three-dimensional solid, we're still integrating on that interval from a to b, but of our cross-sectional area function a(x), whether that's for a square like it was here or whatever other shape we're working with. Then, of course, multiplying by dx. Now, when working with these sorts of problems, we always want to sketch the solid that we're working with as well as a front-facing cross section.

Having these two things will really help you to visualize what's going on in your problem and avoid making any mistakes. Now we're going to continue getting practice with these volume problems coming up in the next couple of videos. For now, you should finish evaluating this integral on your own to get your numerical answer for the volume of this solid. You can check your answer with me in the next video. I'll see you there.

Introduction to Cross Sections Example 1

In the previous video, I showed the full setup of this volume integral based on this three-dimensional shape. So go ahead and check back there if you haven't already before meeting me back here to go through the full evaluation of this integral. For the rest of you, let's go ahead and jump in here. Now, just as a recap, we have our three-dimensional shape here that we formed by taking our curve and extending it upwards. And when we take a slice out of this 3D shape, we were told that these cross sections are squares.

So we set up our volume integral where we're integrating the area function. The area of a square is just that side length squared, and here our side length is \( f(x) \). So we're integrating \( f(x)^2 \). Let's go ahead and finish up evaluating this integral. I'm going to start by squaring this function.

When I do that, this will then become ∫02 16−8x2+x4dx. Then I can find the antiderivative of each of these terms, giving me 16x−83x3+15x5 bounded from zero to two.

Now I can plug those bounds in using the fundamental theorem of calculus. This gives me 16×2−83×23+15×25 and then minus the lower bound plugged in. Each of these terms becomes zero when the lower bound is zero, so we're just subtracting off zero there for the lower bound.

Doing some simplification here, \( 16 \times 2 = 32 \), \( 2^3 = 8 \), \( 8 \times \frac{8}{3} = \frac{64}{3} \), so this is minus \(\frac{64}{3}\), and then \(2^5 = 32\), so that's plus \(\frac{32}{5}\). Now I want to go ahead and combine these fractions by using a common denominator here, giving us \(\frac{256}{15}\). This is just to give all of these that common denominator of 15. Now I can add and subtract all of this across here, which will ultimately give me \(\frac{256}{15}\). If you want to report your answer as a decimal, typing this into a calculator will ultimately give 17.07.

Remember, this isn't just some random number. It actually means something. This represents the volume of this three-dimensional shape if it had square cross sections. Now, if you have any questions here, either on how I set up this integral or on how I ultimately evaluated it, feel free to let us know in the comments below. Thanks for watching. I'll see you in the next one.

Find the volume of the solid whose base is the region bounded by the function $f\left(x\right)=\sqrt{9-x^2}$ and the x-axis with square cross sections perpendicular to the x-axis.

• A. 54
• B. 36
• C. 18
• D. 72

Answer: (B)

Introduction to Cross Sections Example 2

In this problem, we're asked to set up an integral for the volume of a solid that has a base formed by the function y equals four minus x squared on the interval from zero to two. So, coming down to our graph here, the base of all these solids is going to be formed by this area right here, and then we're going to extend it upwards. Now, as we cut into these solids, they're going to have cross sections in different shapes, and these cross sections are going to be cut into parallel to the y-axis. So, we get these cross sections by cutting into our solid parallel to the y-axis. So, looking at part a here, our cross sections are going to be isosceles right triangles.

So, let's go ahead and start here by sketching our solid in a front-facing cross section. I'm going to go ahead and tilt my axis here just to make it a bit easier to draw this looking a little bit three-dimensional. So, if I have my curve on my graph here and I want to go ahead and extend it upward, I can connect those to make this appear three-dimensional. Now, it doesn't matter how good your sketch is as long as it makes sense to you. So, here we have our three-dimensional shape with that base formed by that function y equals four minus x squared.

And when we cut into this parallel to the y-axis, our cross sections are going to be isosceles right triangles. Now, this can be hard to visualize in this three-dimensional shape, so that's why it's important to also go ahead and draw our front-facing cross section. So, when we cut into this and we turn a cross section towards us, it's going to be an isosceles right triangle here. So, if I have my right triangle, that means that I have a 90-degree angle in that corner and being isosceles just means that these two side lengths are equal to each other. So, how are we actually going to find the volume of this solid?

Well, we know that we need to integrate the area function of a cross section. And with our cross section being an isosceles right triangle, the area of a right triangle or any triangle is one-half base times height. So, we need to know what our base and height are in order to create our area function. So, looking back here at my graph, I can see that the base of my triangle is formed by my function four minus x squared. So, the base of my triangle is just f of x, my function.

So, that's the base of my triangle. And because this is an isosceles triangle, this side length is also f of x. So, that means that the area of this triangle is equal to one-half times the base, which is f of x, times the height, which is also f of x. So, ultimately, the area of our cross-section here is equal to one-half times f of x squared. Now we can go ahead and plug in what that function actually is because we know that it's that four minus x squared.

V = ∫ 0 2 1 2 ( 4 - x x2 ) 2 d x

This is our full volume integral. So, this represents the volume of this three-dimensional solid if it had cross sections that were isosceles right triangles.

Find the volume for a solid whose base is the region between the curve $y=\sqrt{\sin x}$ and the x-axis on the interval from $[0,\pi ]$ and whose cross sections are equilateral triangles with bases parallel to the y-axis.

• A. $\frac{\sqrt{3}}{2}$
• B. $-\frac{\sqrt{3}}{4}$
• C. $-\frac{\sqrt{3}}{2}$
• D. $-\frac{4+\sqrt{3}}{4}$

Answer: (A)

Finding Volume Using Disks

We just learned how to find the volume of a three-dimensional solid by integrating the area function of a cross-section taken out of that solid. Now, because these cross-sections could be any shape—a circle, a triangle, a square—determining what that area function was that we were integrating was really the name of the game. Here, we're going to take a look at a particular type of three-dimensional solid whose cross-section is always going to be a circle, making it much easier to determine what that area function is and therefore find the volume. Now here, I'm going to walk you through exactly how this particular type of shape is formed and ultimately how to find its volume. So let's go ahead and get started here.

Now we're going to jump right into our example where we're asked to set up an integral for the volume of a solid formed by rotating the function \( f(x) = 4 - x^2 \) on the interval from negative two to two about the x-axis. Now what exactly does this mean, the solid formed by taking this function and rotating it about the x-axis? Well, we want to take a look at what exactly is happening here by looking at our graph. If we take our function here, \( 4 - x^2 \) plotted from negative two to two, and we take that area and we fully revolve it around that x-axis, we're going to get this three-dimensional shape formed. Now, because this was formed by revolving our curve in a circular motion around that axis, when we cut into our solid, we are going to see these cross-sections that are circles.

So if we take one cross-section out and look at it facing the front, we get this circle with some radius r. This particular type of solid is referred to as a solid of revolution because it was formed by revolving a curve around a particular axis. Now here, that was the x-axis, but it could be the y-axis or some other line, like, say, \( y = 2 \). As we cut into our solid of revolution here on our graph, we can see that these cross-sections end up being circles. And this is true of all of our solids of revolutions. They're always going to have these circular cross-sections that you may hear referred to in this context as discs. Now what's really important here is that because we have these circular cross-sections, our area function is always going to be that of a circle, \( \pi r^2 \). So when we set up our volume integral here, we're going to be taking the definite integral from a to b, whatever interval we're working with, of \( \pi r^2 \), where r is some function of x. So I'm taking that and I'm squaring it and then integrating with respect to x as we have this \( dx \) on the end. Remember that \( dx \) just represents the width of all of our slices as we integrate across that entire solid.

So with this volume integral in mind, what's really important is that we need to figure out what that radius function \( r(x) \) actually is. So with our solids of revolution, our radius is always going to be the distance from the curve to the axis that we revolve that curve around. So coming back down to our solid here, I can see that if I pick any points on the curve of my function and I go down to the axis that we revolve that curve around, that represents my radius. So if I go ahead and take this one cross-section out of my solid here and look at it facing the front, I can see that radius here from my function to that axis point, which in this case is the x-axis where \( y = 0 \). So that means that in order to find my radius function \( r(x) \), I need to take my curve and subtract off my axis.

So I'm just taking \( f(x) \) and subtracting off that axis of revolution, in this case, \( y = 0 \). So that just leaves us with that function all by itself. So, ultimately, our radius function here, \( r(x) \), is just equal to our function \( 4 - x^2 \). So now that we know that radius, we can go ahead and set up that volume integral. Let me go ahead and get out of the way here.

Setting up our volume integral, we are integrating from \( a \) to \( b \). In this case, our interval goes from negative two to two as we integrate up over that whole solid. So from negative two to two of \( \pi r^2 \) where that radius function \( r(x) \) is \( 4 - x^2 \). So \( (4 - x^2)^2 dx \). This integral here represents the volume of this solid of revolution.

If we were to continue in evaluating this here, we would get a numerical answer for this volume. And you should go ahead and try to finish up evaluating this integral on your own, and you can go ahead and check your answer with me in the next video. We're going to continue getting practice with solids of revolution and the disk method coming up next. I'll see you there.

Finding Volume Using Disks Example 3

If you haven't already watched the previous video where I go through the full setup of this problem, go ahead and check it out before coming back to this video where we're going to finish evaluating the integral that we set up there. For the rest of you, let's go ahead and jump into things. Just as a recap, we have this solid that's formed by taking this function from negative two to two and revolving it around our x-axis. Now when we do that, we need to find our volume using the area of these circular cross-sections.

So we are integrating our area function, which is pi r squared, because these are circles. Now our radius is specifically a function of x, where we take our curve and subtract off our axis, which here gave us a radius of four minus x squared, making our whole integral the integral from negative two to two of pi times 4-x22 dx. So now we're going to actually figure out what the volume of this solid of revolution is by fully evaluating this integral. So, let's go ahead and do that here. I'm going to go ahead and pull that pi out front since it's just a constant.

So this is pi times the integral from negative two to two, and then I'm going to go ahead and square this. When I do that, I'll end up getting here 16-8x2 + x4 dx. Now I can go ahead and find my antiderivative here using the power rule. So this gives me pi times 16x-83x3+15x5 then bounded from negative two to two. So now we just need to plug those bounds in using our fundamental theorem of calculus here.

So this is pi times 16×2-8323+1525. That's that upper bound. Then we're subtracting off that lower bound plugged in, which here is negative two. So this is 16×(-2)-83-23+15-25. And that is that lower bound plugged in.

So now we just have to do algebra. We have quite a few fractions going on here, but we're going to wait to deal with those until we do a little bit more simplification. So don't forget about that pi out front, we're keeping it there for now. Then that first term here becomes 32 and then I have a minus two cubed is eight, times eight over three is going to be 64 over three, and then two to the power of five is 32. So this is plus 32 over five, and we're subtracting off here.

16 times negative two, that's negative 32, and then negative two cubed is negative eight times negative eight cubed is going to be positive 64 over three, and then negative two to the power of five. That is going to give us negative 32. So we have minus 32 over five. Now I'm going to go ahead and combine like terms across these two sections with each bound plugged in. I have 32 and then since we're distributing that negative into our parentheses here, I have positive 32 and another positive 32, so that gives me 64.

So this is pi times 64, and then I'm going to go ahead and combine negative 64 over three with another negative 64 over three. That gives us negative 128 over three. And then combining positive 32 over five with another positive 32 over five because we have that double negative. So that is plus 64 over five. Now that we have this down to just three terms, we want to give all of these fractions a common denominator which in this case is going to be 15.

So keeping that pi out front still, multiplying 64 by 15 gives me 960. So I have 960 over 15 then minus 128. If I multiply this by five over five here, 128 times five is 640, so this is minus 640 over 15 and then plus if I multiply 64 over five by three over three, 64 times three gives me 192. So this is 192 over 15. Now I can go ahead and just add all of those numerators across, and this will ultimately give me pi times 512 over 15 which correctly written is going to be 512 pi over 15 for my final answer here.

Now remember, this answer has meaning because this ultimately gives us the volume of our solid of revolution. So here, that volume is 512 pi over 15, and we're all done here. If you have any questions either on how we initially set up this problem or any of the algebra that we did along the way, feel free to let us know in the comments. I'll see you in the next video.

Finding Volume Using Disks Example 4

In this problem, we're asked to set up an interval that represents the volume of a solid of revolution formed by rotating the area between the function \( f(x) = 4 - x^2 \) and the line \( y = 2 \) around that same line \( y = 2 \). So, let's get started here by sketching this solid of revolution. Now we can see our function \( f(x) \) here on our graph. And if we bound that with that line \( y = 2 \) and take a look at this area, this is the area that we're revolving around that line \( y = 2 \) to form our solid. So, if I go ahead and sketch the other half of this solid of revolution, we can see that it forms this sort of three-dimensional oval-type shape.

So, this is the solid that we want to find the volume of. Now, looking at this solid, if I go ahead and take a cross-section out of it, if I slice it down the middle and then have a front-facing cross-section, its cross-section is going to be a circle. And we know that the radius of our cross-section goes from our curve, in this case, \( f(x) \), to our axis, which here is at \( y = 2 \). So in order to set up our volume integral, there are two things that we need to find here. We need to find our radius function, \( r(x) \), and we also need to determine the bounds of our integral.

So, let's start with our radius function. We know that \( r(x) \), our radius, is going to be our curve minus our axis. Here, our curve \( f(x) = 4 - x^2 \), and then our axis is at \( y = 2 \). So it's important to note here that we are not revolving this around the x-axis where \( y = 0 \). We're actually revolving it around a different line, \( y = 2 \).

So, that means that we're going to subtract off two here. So when we do that, combining our like terms four and negative two, this ultimately gives us \( 2 - x^2 \) for our radius function. So now that we have this radius function, the only other thing we need to set up our integral is the bounds. So, coming back over here to our graph, what values of x bound my solid here? Now I know that based on where my function intersects that line, \( y = 2 \), these are going to represent the bounds of my integral.

We need to figure out what these x values are. And we can do that by setting our function equal to \( y = 2 \). This is how we find intersection points. So, if we take \( 4 - x^2 \) and set that equal to \( y = 2 \), we can go ahead and subtract four on both sides to cancel that. Then we have \( -x^2 = -2 \).

Canceling those negatives, we just have \( x^2 = 2 \). Then, if I take the square root on both sides, this ultimately gives me that \( x \) is equal to plus or minus the square root of two. So these intersection points are where \( x = -\sqrt{2} \) and where \( x = \sqrt{2} \). These are the bounds of my integral. So, with that out of the way, we can go ahead and set up our volume integral.

So, our volume is going to be equal to the integral from -√2 to √2 of \( \pi r^2 \), where \( r \) is that function \( 2 - x^2 \). So, this is \( \pi \times (2 - x^2)^2 \, dx \). And this integral ultimately represents the volume of our solid of revolution. So if we were to continue in evaluating this integral, we would get a numerical answer that is ultimately the volume of this solid.

Thanks for watching, and I'll see you in the next one.

Find the volume of the solid obtained by rotating the region bounded by $y=x+4$, $y=0$, $x=1$ & $x=5$.

• A. $\frac{854}{3}\pi$
• B. $4\pi$
• C. $\frac{604}{3}\pi$
• D. $604\pi$

Answer: (C)

Disk Method Using y-Axis

We just learned how to find the volume of a solid of revolution with circular cross sections using the disk method, and we found that a solid of revolution is formed by taking a function and revolving it around a particular axis, whether that's the x-axis, the y-axis, or some other line. Now we focus specifically on setting up our volume integral for a function revolved around the x-axis, and we're going to follow a super similar process when we now take a function and revolve it around the y-axis. But our math has to change slightly because our radius function inside of that integral can no longer be a function of x as it was when we revolved around the x-axis, but must now be a function of y as we revolve around the y-axis. Now I'm going to fully break this down for you as we work through our example together. So let's go ahead and jump in here.

Here we're asked to set up an integral for the volume of a solid formed by rotating the function f(x) = 4 - x2 on the interval from zero to two about the y-axis. So coming down to our graph here, we can see our function 4 minus x2 on the interval from zero to two. So we're taking that function curve and fully revolving it around the y-axis in order to get this solid that's sort of shaped like a dome. Now we want to find the volume of this solid, so how exactly can we set this up? Well, when we took this same function and instead revolved it around the x-axis, we got our volume interval by integrating πr2, that area function for those circular cross sections.

Now in doing that, our radius was a function of x, and we therefore integrated with respect to x as we evolved around the x-axis. So we're going to set up a super similar integral here. We are still integrating πr2, but now because we're evolving around the y-axis, our radius must now be a function of y, and we therefore have to integrate with respect to y. So before we can fully set up this integral, let's break this down a bit further. Whenever we're revolving around the y-axis, the very first thing we want to do is write our function in terms of y.

Now this is because we are looking at our function in relation to our y-axis as opposed to our x-axis. So we want to take this function f(x) = 4 - x2 and get it in terms of y by solving for x. So if f(x) = 4 - x2, that's the same thing as saying y = 4 - x2. So I can get this in terms of y by solving for x. So I am going to add x2 to both sides here to get that to cancel and then also subtract y from both sides.

So I am left here with x2 = 4 - y. Now I can go ahead and fully isolate x here by square rooting both sides to cancel that squared. So I get here that x = 4 - y. Now I only care about the positive square root here because we're only working in that first quadrant with our function. So now that I have solved for x, I have this function of y that we can now call g(y).

So I've rewritten my function in terms of y. So now let's take this a step further. The next thing we want to do is determine how exactly we're going to set up this integral by finding that radius function r(y) and the bounds of our integral, which must also be values of y. So let's go ahead and take a closer look at this circular cross-section. So if I take this cross section and I want to look at it facing the front, I know that my radius goes from that center point, which is my y-axis, out to my function curve.

Now having done this before, we know that our radius is going to be our curve minus our axis because it's that distance from our curve to that axis. Now in this case, our curve is that function of y, g(y), that we just found up above, and we have that axis at the center, our y-axis, where x is equal to zero. So in order to find our radius function here as a function of y, we're taking our curve g(y) and subtracting off our axis, which is just zero. So if I take g(y) minus zero, that just leaves me with g(y) itself, which I know to be just the square root of four minus y that we found up above. So now I have my radius function r(y) that I'll eventually put into my integral.

Now the other thing that

Disk Method Using y-Axis Example 5

Here we're asked to find the volume of a solid formed by rotating the function f(x)=4−x2 on the interval from zero to two about the y-axis. Now if you haven't already watched the previous video where I show exactly how to set up this integral, go ahead and go back there before continuing on with me here. Just as a recap, remember that we took our function and rewrote it in terms of y. That allowed us to find our radius, which is a function of y, and we also found the bounds of our integral, y values, and we put all of that into our volume integral here where we're integrating πr2, that area function of our circular cross-sections. So now that we actually have this volume integral, we can just go ahead and fully evaluate it to find that numerical answer and ultimately the volume of this solid.

Let's continue evaluating our integral here. We have the integral from zero to four of π×4−y2dy. We can see that the squared and the square root are going to cancel out nicely there, and I can go ahead and pull that π out front since it's just a constant and rewrite this as the integral of π times the integral from zero to four of four minus y dy. So now that I have this integral, I can go ahead and find that antiderivative here, keeping that constant π out front, applying my power rule here. This is going to be 4y−12y2 and then bounded from zero to four.

Now we can just go ahead and apply those bounds. Plugging in my upper bound here, this is π times four times four minus one half four squared. When we plug zero into both of these terms, because they have a y multiplying in both cases, both of those terms are going to end up being zero. So we're really just subtracting off zero here. Simplifying this, four times four is 16.

This is π×16−1242 that's going to be eight, and then 16 minus eight is just eight. So I get a final answer here of 8π for the volume of this dome-shaped solid. If you have any questions on how exactly I set this up or got to this 8π, feel free to leave a question in the comments. I'll see you in the next video.

Disk Method Using y-Axis Example 6

In this problem, we're asked to find the volume of the solid of revolution formed by rotating the area between the function f(x) = 4 - x², the line y = 4, and the line x = 2, specifically only in quadrant one around that line x = 2. So let's start here by just sketching our solid of revolution and one of its cross sections. We know that we're only working in the first quadrant and looking at the area between our function's curve, the line y = 4, and the line x = 2. And we're taking this area and revolving it around the line x = 2 to form our solid.

I'm going to go ahead and sketch the other side of this solid since we're revolving it around that line. And ultimately, this forms that three-dimensional solid of revolution that I'm going to shade in yellow here. We have this sort of funnel-like shape that we're dealing with. And I also want to go ahead and slice into this solid and draw one of its cross sections. These have circular cross sections since they are solids of revolution.

With our solid and its cross section sketched, let's talk about how we're actually going to set up our volume integral. We know that we need to be integrating the cross-sectional area function, which since our cross sections are circles, is πr². Since we are revolving this area around a vertical line like the y-axis, that ultimately means that our radius needs to be a function of y, and we need to integrate with respect to y. That also means that the bounds of our integral should be y values.

Thinking first about what that radius should be, looking at our circular cross section, we know that our radius is the distance from our curve to our axis. Our axis of revolution here is the line x = 2. If our radius needs to be a function of y, that means that our curve is also a function of y. We need to take our original function and rewrite it to get x in terms of y so that we can then use that to find our radius function. If y = 4 - x², by rearranging this, we find x² = 4 - y, and taking the square root on both sides, we get x = √(4 - y). We only care about the positive square root since we're only working in the first quadrant. This is our curve as a function of y.

Now, we use this to find our radius. The radius of our circular cross section is the difference between our curve and our axis of revolution; thus, r = √(4 - y) - 2. It's crucial to recognize that the axis of revolution here is the line x = 2.

Now to determine the bounds of our integral, which should be y values since we are integrating with respect to y. We're working in the first quadrant, bounded by the line y = 4, thus the upper bound is 4, and the lower bound is 0. Setting up our volume integral, we bound it from 0 to 4. The integrand is πr², where r = √(4 - y) - 2, squared and integrated with respect to y.

After setting up the integral, we evaluate it. First, we square the function: integral from 0 to 4 of π(8 - y - 4√(4 - y)) dy. Distribute the π, split into multiple integrals if needed, and solve each separately. After using u-substitution and simplifications, you find the integral values, sum them, and obtain the final answer. Remember, it represents the volume of the solid of revolution formed in the problem.

Find the volume of the solid formed by revolving the area bounded by $f\left(x\right)=x^2$ from $x=0$ to $x=3$ and the y-axis around the y-axis.

• A. $81\pi$
• B. $162\pi$
• C. $\frac{81}{2}\pi$
• D. $324\pi$

Answer: (C)

Here’s what students ask on this topic:

What is the disk method in calculus for finding volume?

The disk method is a technique used in calculus to find the volume of a solid of revolution. This method involves revolving a function around an axis, typically the x-axis or y-axis, to create a three-dimensional shape. The cross-sections of this solid are circular, resembling disks. To calculate the volume, you integrate the area of these circular cross-sections, which is given by the formula $\pi r^2$, where $r$ is the radius of the disk. The radius is determined by the distance from the curve to the axis of revolution. The integral is set up over the interval of the function, and the result gives the total volume of the solid.

How do you find the volume of a solid with square cross sections?

To find the volume of a solid with square cross sections, you first need to determine the area of one square cross section. If the base of the solid is defined by a function $f\left(x\right)$, the side length of the square is $f\left(x\right)$. Therefore, the area of the square is $f^{\left(}$. To find the volume, integrate this area function over the interval of the base. The integral is set up as ${\int }_a^b{f}^{\left(}dx$, where $a$ and $b$ are the bounds of the interval. The result of this integral gives the total volume of the solid.

What is a solid of revolution in calculus?

A solid of revolution is a three-dimensional shape created by rotating a two-dimensional curve around a specific axis. This process generates a symmetrical solid, where the cross-sections perpendicular to the axis of rotation are circular. The volume of such a solid can be calculated using methods like the disk or washer method, depending on the specific setup of the problem. The disk method involves integrating the area of circular cross-sections, while the washer method is used when there is a hollow region within the solid. These methods are essential in calculus for solving problems related to volumes of solids formed by rotation.

How do you set up an integral for the volume of a solid of revolution?

To set up an integral for the volume of a solid of revolution, first identify the axis of rotation and the function being revolved. If revolving around the x-axis, the radius of the circular cross-section is the distance from the curve to the axis, given by the function $f\left(x\right)$. The area of the cross-section is $\pi f^{\left(}$. Integrate this area over the interval of the function: ${\int }_a^b\pi f^{\left(}dx$. If revolving around the y-axis, express the function in terms of $y$ and integrate with respect to $y$. The integral provides the volume of the solid.

What is the difference between the disk method and the washer method?

The disk method and the washer method are both techniques used to find the volume of solids of revolution, but they differ in their application. The disk method is used when the solid has no hollow region, meaning the cross-sections are solid disks. The volume is calculated by integrating the area of these disks, $\pi r^2$, where $r$ is the radius. The washer method, on the other hand, is used when there is a hollow region within the solid, resulting in washer-shaped cross-sections. The volume is found by integrating the difference between the outer and inner radii, $\pi (R^2-r^2)$, where $R$ is the outer radius and $r$ is the inner radius.