9-34. Shell method Let R be the region bounded by the followin...

Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

y = x³−x⁸+1,y=1; about the y-axis

Accepted Answer

Hello. In this video, we are going to let R be the region bounded by Y is equal to x 4th minus X 9 + 3, and Y is equal to 3. We want to find the volume of the solid generated when R is revolved about the y axis using the shell method. Now, there are a few things that we can do here, but let's go ahead and define the integral that is going to give us our volume. Now, that integral is going to be V is equal to 2 pi, multiplied by a definite integral from A to B, of our radius. Multiplied by the height of the shell. And because we are integrating or rotating about the y axis, we are going to integrate with respect to X. Now, what we need to do is we need to define our radius and our height. For the given shell. So let's go ahead and draw a shell that is oriented about the X axis. Now, starting with the radius from the center of the shell to the edge of the shell, because this shell is oriented about the X axis, that means that the radius is going to be some measure of X. So we are going to label radius as the variable X. Now, what about the height of the shell? Now, there are two equations that are given to us in the problem. We are given Y is equal to X the 4th, minus X 9th plus 3, and Y is equal to 3. Now, here, height is going to be defined. As our uppermost curve minus our lowermost curve. Now, in this case here, if we take a look at the first equation, this first equation has a higher magnitude when compared to the second equation. So what this means is that the uppermost part of our shell is going to be defined as X 4th power minus X 9 + 3, and our lower part of the shell is going to be defined as Y is equal to 3. What this means is that the height is going to be defined as x 4th power minus x 9th plus 3 minus the lower curve, which is the constant value of 3, and this is going to simplify to leave us with X to the 4th power minus X 9th power. So this is going to be the value of the height. Now what we need to do is we need to find the bounds of integration. Now, in order to find the bounds of integration, we want to find the values of X where the two equations intercept each other. In order to do this, we will take both of our equations, set them equal to each other, and solve for X. So that is going to give us X to the 4th power minus X 9th, plus 3, and that is equal to 3. We can start by subtracting 3 to both sides of the equation, leaving us with X to the 4th, minus X the 9th equal to 0. Next, if we factor out an X of the 4th from the left hand side of the equation, we get X of the 4th power multiplied by 1 minus X to the 5th power. And by setting both of these factors of X equal to 0 and solving, we get X's equal to 0 and X is equal to 1. So what this means is there are bounds of integration are going to be from 0 to 1. So now that we've defined our radius, our height, and the bounds of integration, let's go ahead and construct the volume integral. So, our volume integral is going to be 2 pi, multiplied by the integral from 0 to 1 of our radius X, multiplied by our height, which is X to the 4th power minus X 9th, and we will be integrating with respect to X. All we have to do at this point is we need to solve the integral in order to find the volume. So, the first thing we will do is we will distribute X into the integrand. That will give us 2 pi multiplied by the integral from 0 to 1 of X to the 5th power, minus X 10th power DX. Then by taking the antiderivative, we will be left with 2 pi multiplied by 1/6 X 6 power, minus 1 divided by 11th, multiplied by X 11th power, and this is going to be evaluated from 0 to 1. By using the fundamental theorem calculus part two, we will go ahead and plug in our upper bound first, and that is going to leave us with 1/6 minus 1/11. Then we will plug in the lower bound and subtract that value, but if we plug in zero, we are just left with 0. So this is going to simplify to be 2 multiplied by 1/6 minus 1/11. Then by combining like terms, this is going to give us 2 pi multiplied by 5 divided by 66. And by multiplying the current terms together, we get 10 pi divided by 66, which will further simplify to be 5 pi divided by 33. This is going to be the value of the volume of the revolved shape. And with that being said, the overall solution to this problem is going to be C. So I hope this video helps you in understanding how to approach the problem, and we will go ahead and see you all in the next video.

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis. y = x³−x⁸+1,y=1; about the y-axis

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9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

y = x³−x⁸+1,y=1; about the y-axis