69. Comparing volumes Let R be the region bounded by y = sin x...

Question

69. Comparing volumes Let R be the region bounded by y = sin x and the x-axis on the interval [0, π]. Which is greater, the volume when R is revolved about the x-axis, or the volume when R is revolved about the y-axis?

Accepted Answer

Hello, in this video, we are going to let T be the region bounded by Y is equal to 2X minus X squared, and the X axis on the interval from 0 to 2. We want to determine which of these is greater. Is the volume going to be greater when we revolve T about the X axis, or when we revolve T about the Y axis? So, in order to approach this problem, we are going to have to find the volume for both the shapes rotated about the X and Y axis. Now, in this case here, let's go ahead and start by determining the volume when the shape is rotated about the X axis. Now, if we were to rotate the shape about the X-axis, we will end up with a symmetric shape with no gaps in between. And what this means is that if we rotate the X-axis, we can determine the volume by using the disk method. Now, if we are using the disk method, the volume is going to be defined as the following. Volume is equal to pi, multiplied by a definite integral from A to B, of the radius of the disk squared DX. Now, in this case here, the radius is defined as the distance from the X-axis to the uppermost curve, and in this case here, the curve is defined as 2 X minus X squared. So radius in this case is going to be 2 X minus X2. Now, we need to determine the bounds of integration, but in this case here, that is given to us in the problem. We want to integrate from 0 to 2 on the X-axis. So, by taking these pieces of information, volume for the X axis is going to be defined as pi multiplied by the integral from 0 to 2 of our radius squared, which is 2X minus X2, quantity squared X. Now what we are going to do is we are going to solve for this volume. Now, the first thing we're going to do is we are going to expand the integrand. That is going to leave us with pi multiplied by the integral from 0 to 2 of the quantity 4 X squad. Minus 4 X squared. Sorry, -4 X cubed. Plus x to the 4th power d x. Then by taking the antiderivative of each of these terms, we will be left with pi multiplied by 4/3 x cubed minus x plus 1/5 x 5th power. And this will be evaluated from 0 to 2. Then by using the fundamental theorem of calculus part 2. We will be left with pi multiplied by the values of 32 divided by 3 minus 16 plus 32 divided by 5 minus 0, and by combining the like terms together, we will have a final volume of 16 pi divided by 15. And so this is going to be the volume when we are rotating about the X axis. So, we are going to take this value and set it to the side for now. Now we need to determine the volume when we rotate the shape about the Y axis. Now, in this case here, if we were to rotate about the Y axis, we will end up with a symmetric shape. Where the curve overlaps each other by both the disk and the washer method. And so in order to approach this kind of problem, we are going to use the shell method. Now, if we use the method of cylindrical shells, the volume is going to be defined as two pi multiplied by a definite integral from A to B. Of the radius of the shell, multiplied by its height, and in this case here, because we're rotating about the Y axis, we are going to integrate with respect to X. Now, these boundaries are going to be the same as the first integral we worked on. The boundary along the X axis is going to be from 0 to 2. Now, the radius of the shell is just going to be some measure of X, so radius is going to equal to the variable X. And again, the height is going to be defined as the distance from the X axis to the uppermost curve, and that is going to be 2X minus X squared. And so by taking these pieces of information, when we rotate about the Y axis, the volume will be defined as 2 pi multiplied by the integral from 0 to 2, of X multiplied by the quantity. 2 X minus X 2 DX. And now we are going to solve this integral to get the volume. So by distributing X, we can rewrite the integral to be 2 X2 minus X cubed DX. Then when we take the antiderivative, we will be left with 2 pi multiplied by the quantity, 2/3 X cubed minus 1/4, 4th power, and this will be evaluated from 0 to 2. Now, if we go ahead and apply the fundamental theorem of calculus part 2, we'll be left with 2 pi, multiplied by the quantity. 163. -4, 0. And by combining the like terms together, we will get a volume of 8 pi divided by 3. This is going to be the volume of the shape when we revolve it about the y axis. So now that we have both of the volumes, when we revolve about the X and Y axis, we are going to compare these volumes together. So, we are comparing 8 pi divided by 3 to 16 pi divided by 15, but in this case here, 8 pi divided by 3 is going to be the larger quantity. And so with that being said, we get a greater volume when we integrate by rotating about the y axis in comparison to rotating about the X axis, which is going to be the solution to this problem. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

69. Comparing volumes Let R be the region bounded by y = sin x and the x-axis on the interval [0, π]. Which is greater, the volume when R is revolved about the x-axis, or the volume when R is revolved about the y-axis?

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