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

this problem, we're asked to find the volume of the solid obtained by rotating the region bounded by y equals x plus four, y equals zero, x equals one, and x equals five about the x axis. Now let's start here by sketching what this solid looks like. Now we're gonna start here by just looking at that region bounded by these four lines here. Now we're starting with y equals x plus four. I'm just gonna sketch this line. We know that that gives us a y intercept at four and then just a sloped line. Then the line y equals zero, that's just the x axis that I have with this dotted line. So that's y equals zero. And we also have our lines x equals one and x equals five. So I'm gonna go ahead and sketch those using a dotted lines as well. So this is the line x equals one, and then we also have the line x equals five. Now, if we look at all of those restrictions, if we look at this bounded by x plus four, the line y equals zero, the line x equals one, and the line x equals five, we're gonna take this region and we're going to rotate it around the x axis. So I'm gonna go ahead and sketch the other half of this solid by sort of mirroring that side over there and then revolving this around that x axis here, drawing this dotted line just to show that this is a three-dimensional shape. So we get this sort of sideways plateau looking shape here, and we want to find the volume of the shape. So how exactly can we do that? Well, since this is a solid of revolution and we formed it formed it by taking our area, revolving it around our x axis, when we take a slice out of this, we can see that our cross sectional areas or our cross sections are these circles. So these circles have some particular radius that's formed from our curve, which is f of x, to our axis, which is our x axis where y equals zero. So to find the volume here, we need to integrate that cross sectional area Pi r squared since these are circles. So we want to go ahead and find that radius r of x here. So r of x is gonna be equal to my curve minus my axis here. That's f of x minus my x axis, which is just zero. So this ultimately gives me just x plus four, just that function itself. So this is my radius function, and I wanna put this into my volume integral. Now when I set up my volume integral, I need to bound it. So what should the bounds of my integral be? Well, looking at my graph here, I know that these need to be x values since we're integrating with respect to x. And I can see that this solid starts at x equals one, ends where x equals five based on what was given to us in the problem. So those are the bounds of my integral, one to five. Then we're integrating pi r squared where r is that x plus four, and then integrating with respect to x. So with this volume integral set up, we can go ahead and evaluate it. And there are a couple different ways you could go about evaluating this integral. We could go ahead and just multiply x plus four out by itself to square it, or you can do a u substitution here. It is up to you. You will get the same answer regardless. I'm gonna go ahead and show you how I would do this using a u substitution. So first thing I'm gonna do is pull that pi out front. So this is the this is pi times the integral from one to five of x plus four squared DX. Then if we set u equal to x plus four, that then makes DU just equal to DX. So in my integral, I have u and I have du. Now in doing this, I'm gonna go ahead and also transform the bounds of my integral along with the substitution so that I don't have to put this original function of x back in later. So I'm gonna go ahead and find my new balance here by plugging one and five in for x for my choice of u. So this is one plus four and five plus four. So that gives me new balance here of five and nine. So my new integral here with this substitution, remember, I still need to keep that pi out front, is the integral from five to nine, those new bounds of u squared du. So whenever I take this integral, this is PI times one third, U cubed, bounded from five to nine. Now I'm gonna go ahead and put those constants out front together, PI over three, and then apply those bounds. So this is going to be nine cubed minus five cubed. Now if I take nine cubed minus five cubed, that will ultimately give me an answer here of six zero four Pi over three because remember, we want to put that constant ultimately in front of that Pi. So this is my final answer here, 604 pi over three. And remember this isn't just a random number it actually represents something. This is the volume of this solid of revolution. Now if you have any questions here as to how I set this up or any of the algebra that I did along the way, feel free to leave us a comment below. Thanks for watching, and 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 obtained by rotating the region bounded by $y = x + 4$ , $y equals 0$ , $x equals 1$ & $x equals 5$ about the x-axis.

$854 over 3 pi$

$4 pi$

$604 over 3 pi$

$604 pi$

Verified step by step guidance

Identify the region to be rotated: The region is bounded by the lines y = x + 4, y = 0, x = 1, and x = 5.

Determine the axis of rotation: The problem does not specify, but typically such problems involve rotation around the x-axis.

Set up the integral for the volume of the solid of revolution using the disk method: The formula is $ V = \int_{a}^{b} \pi [R(x)]^2 \, dx $, where $ R(x) $ is the distance from the curve to the axis of rotation.

For this problem, $ R(x) = x + 4 $ since the region is rotated around the x-axis. The limits of integration are from x = 1 to x = 5.

Substitute $ R(x) $ and the limits into the integral: $ V = \int_{1}^{5} \pi (x + 4)^2 \, dx $. Expand the integrand and integrate term by term to find the volume.