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

this problem, we're asked to find the volume of the solid formed by revolving the area bounded by f of x equals x squared from x equals zero to x equals three and the y axis, specifically revolving this around that y axis. So let's start here by just sketching our solid. Now we know we're dealing with a function f of x equals x squared from x equals zero to x equals three. So I'm gonna go ahead and plot that curve right there. Now I know that when x is equal to three, that corresponds then to a y value of nine. So we're looking specifically at this area that I'm going to go ahead and shade in yellow here, and we're taking that area and we are revolving it around our y axis. So if I go ahead and sketch the solid that that forms here, we get a sort of bowl shape. So this is my three-dimensional bowl shape. Now it's okay if your sketches aren't perfect. Mine definitely are not as long as it helps you visualize what's going on in the problem. So now we have this sort of three d bowl shape. Now looking at this cross section, we have our circular cross section here because this is a solid of revolution, and we know that this circular cross section has some radius y that goes from our curve to our axis of revolution in the middle there. So with that sketched, let's think about what our volume integral should look like. Now we know here since we're revolving around the y axis that our radius needs to be a function of y and we're integrating with respect to y, which means that the balance of our integral should also be y values. Now looking at this here, since we're dealing with the function y equals x squared, the first thing I want to do here is rearrange this to get x in terms of y. And I can easily do that by taking the square root on both sides. If I then flip which side this is on, I get that x is equal to the square root of y. And I only care about that positive square root because we're only looking at the portion of our curve in that first quadrant from x equals zero to x equals three, and that's what we were evolved around. So this x equals root y represents my function in terms of y. So looking at my circular cross section here, I can then say that this radius goes from my curve G of y to that axis point at the middle, which is where x equals zero. That's the y axis. So I know that my radius here, r of y, is then equal to my curve minus my axis, which here is root y minus zero, which means that it's just root y. So with that radius found, we just need to determine the bounds of our integral and then we can go ahead and set it up. Now looking at the sketch that we did here, we know that this is bounded from y equals zero up to y equals nine. Because this is bounded from this is on the interval from x equals zero to x equals three. If we were to plug zero and three into our function f of x, that gives us a zero and nine for our corresponding y values. So when setting up our volume integral here, this is then equal to the integral from zero to nine of pi r squared, where r is the square root of y dy. So now we just need to evaluate this integral. Now this works out rather nicely because our squared cancels with our square root, leaving us with just the integral from zero to nine of pi y d y. Now because pi is just a constant, I can pull that out front. Further simplifying this to Pi times the integral from zero to nine of y dy. Then I can go ahead and use the power rule here. This gives me Pi times one half y squared bounded from zero to nine. I'm going to go ahead and combine those constants to just be pi over two and then plugging my bounds into that y squared, that's nine squared minus zero squared. Zero squared is just zero and nine squared is 81, so this becomes 81 pi over two and this is my final answer here which gives me the volume of this solid of revolution. Now if you have any questions here either about the setup 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 one.

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

Calculus

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

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Multiple Choice

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

$81 pi$

$162 pi$

$81 over 2 pi$

$324 pi$

Verified step by step guidance

To find the volume of the solid formed by revolving the area bounded by f(x) = x^2 from x = 0 to x = 3 around the y-axis, we use the formula for the volume of a solid of revolution: $ V = \int_{c}^{d} \pi [R(y)]^2 \, dy $.

First, express x in terms of y from the function f(x) = x^2. Since y = x^2, we have x = \sqrt{y}.

Determine the limits of integration. The function f(x) = x^2 ranges from x = 0 to x = 3, which corresponds to y = 0 to y = 9.

The radius of the solid at any point y is given by R(y) = x = \sqrt{y}. Substitute this into the volume formula: $ V = \int_{0}^{9} \pi [\sqrt{y}]^2 \, dy $.

Simplify the integrand: $ [\sqrt{y}]^2 = y $. Therefore, the integral becomes $ V = \pi \int_{0}^{9} y \, dy $. Evaluate this integral to find the volume.