For the following regions R, determine which is greater—the vo...

Question

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.

R is bounded by y=1−x^3, the x-axis, and the y-axis.

Accepted Answer

Hello. In this video, we are going to let R be the region bounded by Y is equal to 3 minus X2, the X-axis, and the Y axis. When R is revolved about the X-axis and about the y axis, which solid has the greater volume? We are going to approach this problem by starting with the graph or the shape of the graph. So, by drawing an axis and graphing Y is equal to 3 minus X2, the curve given to us is a negative parabola, whose vertex is located at 3 on the Y axis and opens negatively. This is going to be the shape of the equation Y is equal to 3 minus X2. Now, we know that this region is also bounded by the X axis. And is bounded by the y axis. Now, before we decide to find the volumes, let's go ahead and find the X intercepts of this parabola. In order to do this, we are going to set the parabola equal to 0 and sulfur X. That gives us 3 minus X2 is equal to 0. Adding X2 to both sides gives us X2 equal to 3, and by taking the square root of both sides of the equation, we get X's equal to positive and negative square root of 3. So this parabola has intercepts located at X is equal to square root of 3, and X is equal to negative square root of 3. Now this region R is going to be the entirety of the shaded region. But now the question is, which revolve when we revolve this shape, which is going to produce a higher volume, revolving about the X axis or revolving about the Y axis. So in order to answer this question, we are going to need to find the independent volumes. Let's go ahead and start by revolving the shape about the Y axis. Now, if we were to revolve the shape about the Y axis, we are going to have to integrate with respect to X. Notice that when we revolve about the y axis, we have a solid shape which is revolved. That means that we can define this volume by using the disk method. So the volume with respect to X is going to be defined as pi multiplied by the integral from A to B. Of a radius squared. DX Now, our radius of the shape is going to be defined as the parabola. The parabola is going to be 3 minus X2, which is already in terms of X. But what about the boundaries? Well, the boundaries are going to be from -3, 2 root of 3, to square root of 3, so the volume with respect to X is going to be pi multiplied by the integral from negative square root of 3. To square root of 3 of the quantity, 3 minus X2 quantity square DX. Now notice one more thing. If we were to revolve the shape about the Y axis, notice that we have a positive symmetric shape. So we can further simplify this integral by taking twice the area from 0 to square root of 3. So we can rewrite this volume in terms of X one more time, as 2 pi multiplied by the integral, from 0 to square root of 3. Of 3 minus X2 quantity squared X. Now let's go ahead and solve for this volume. The first thing we're going to do is we're going to expand the integrand. That is going to leave us with 2 pi multiplied by the integral from 0 to square root of 3 of 9 minus 6 X plus X6X squad plus X to the power of 4DX. And by taking the antiderivative of both sides of this equation, We get 2 pi multiplied by the quantity, 9 X minus 2 X cubed plus 15th 5th power, and this will be evaluated from 0 to square root of 3. Now what we need to do is we need to plug in both of the boundaries. Now, plugging in square root of 3 is going to leave us with 2 pi multiplied by the integral, 9 multiplied by the square root of 3. Minus 2 multiplied by the square root of 3, quantity cued plus 15th multiplied by square root of 3, raise to the 5th power. Then we need to subtract the value of the lower boundary, but if we plug in the lower boundary 0, we are just left with 0. Now, what we are going to do is we are going to use a calculator to help us approximate the value, and the approximate value of this volume is going to be 26.1 i. So, when we revolve about the Y axis and integrate the volume with respect to X, that is going to give us a volume of 26.1 pi. Now we need to repeat this process, but we are going to revolve about the X axis. Now, when we revolve about the X-axis, we are going to have to integrate with respect to Y. The smallest value of Y in the graph is going to be Y is equal to 0, and the largest value of Y is going to be at Y is equal to 3. So when we revolve about the X-axis, that is going to give us V of Y on the bounds from A to B, which is 0 to 3. So we have i multiplied by the integral from 0 to 3. But what about the radius? Well, the radius is going to be the parabola, but we need to represent the parabola in terms of Y. If we were to solve for X and express it in terms of Y, then we get the integrand to be the square root of 3 minus Y. This quantity will be squared D Y. Now what we need to do is we just need to solve this integral. The integrand will simplify and leave us with pi multiply with the integral of 0 to 3 of 3 minus YDY. This will further simplify to be pi multiplied by 3 Y minus 12 Y2d, evaluated from 0 to 3. By plugging in our upper boundary, we get pi multiplied by 3, multiplied by 3, which is 9, minus 9 divided by 2, and if we plug in 0, we just get 0. So, by further simplifying and using a calculator, this is going to give us 4.5 pi. And so what this means is that if we revolve around the X axis and integrate with respect to Y, this is going to give us a volume of 4.5 pi. But if we compare the two volumes together, we know that the volume with respect to X is greater than the volume with respect to Y, meaning that when we revolve the shape about the Y axis, we get the larger volume. And with that being said, the overall solution to this problem is going to be A. So I hope this video helps in understanding how to approach this problem, and we will go ahead and see you all in the next video.

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.R is bounded by y=1−x^3, the x-axis, and the y-axis.

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For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.

R is bounded by y=1−x^3, the x-axis, and the y-axis.