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.

{Use of Tech} y = √sin^−1x,y = √π/2, and x=0; about the x-axis

Accepted Answer

Hello. In this video we are going to let Q be the region bounded by Y is equal to the square root of sine inverse of 2 X. Y is equal to the square root of pi divided by 6, and X is equal to 0. We want to find the volume of the solid generated when Q is revolved about the x axis using the shell method. Now, there are a few things that are labeled on the given graph. First, this green curve represents the sine function, and this blue curve represents the constant function of Y. Now there is a third curve that is given to us, and that is X is equal to 0. X is equal to 0. It is a vertical line that overlaps the y axis. So, this orange region is going to be the bounded region Q. Now, two more things that the problem tells us is that we are revolving the shape about the X axis. And more importantly, we want to find the volume by using the shell method. Now, since we are using the Shall method, and we are revolving this region about the X-axis, we are going to find the volume by integrating with respect to Y. So volume is going to be defined as 2 pi, multiplied by a definite integral from A to B, of the radius of the shell, multiplied by its height, and this will be integrated in terms of Y. So, what we want to go ahead and do is make sure that our giving curves are X in terms of Y. And so what we'll need to do with this green curve is we'll need to go ahead and solve for X in terms of Y. Now, the first thing we can do is we can square both sides of this equation. That will leave us with sine inverse of 2 X equal to Y2d. Then by taking the normal sine function of both sides of this equation, we will have 2 X is equal to sine of Y squared. And finally, to isolate X on the left hand side, we will divide both sides of this equation by 2, and that will leave us with X's equal to 1/2 sin of Y squared. So this is going to be the equation of the green curve in terms of X in terms of Y. So, let's go ahead and start by defining the bounds of integration. Now, since we are integrating with respect to Y, we want to first identify the lowest value of Y. Now in the given region. The lowest possible value of Y is going to be located at the origin where Y is equal to 0, so the lowest boundary is going to be 0. Now, the largest boundary is going to be determined by the intersection of the blue and green curve, and that is going to be given as this constant function of Y. So, the largest value of Y is going to be the square root of pi divided by 6. So this is going to be your bounds of integration. Now, what about the radius? Now, because we are integrating with respect to Y, the radius is just going to be some measure of Y, so radius will be the variable Y. Now, what about the height of the equation? While the height of the equation is going to be defined as the rightmost curve minus the leftmost curve. Now, in this case here, the rightmost curve is defined by the green curve, and we've already solved for that curve as X in terms of Y. So that is going to be one half of Y2 minus the smallest value of Y, which in this case is going to be 0. And so that means our height is going to be defined as 1/2. Sign of Y squared. And so now that we have all the pieces we need, we can go ahead and construct the integral that will give us the volume for this region. So the volume is going to be defined as 2 pi multiplied by the integral from 0 to the square root of pi divided by 6, and this will be multiplied by our radius Y. Multiplied by our height, which is 1/2 of Y 2 D Y. And now what we need to do is we need to solve for this integral in order to find the volume of the region when it is rotated about the X-axis. Now, the first thing we're going to do is we're going to remove this one half to the front of the integral as a constant, and that is going to allow us to rewrite the volume as pi, multiplied by the integral from 0 to the square root of pi divided by 6, of Y multiplied by sine of Y squared D Y. Now, the next thing we're going to do is we're going to integrate this by using a use substitution. So we're going to allow you. To equal to squared, then DU will equal to 2 Y D Y. Now, we have a YDY inside of the integral already, so we just need to remove this two coefficient from the DU statement. By dividing both sides of this equation by 2, we will have 12 DU is equal to YDY. And so when we apply this use substitution, we will have the constant one half multiplied by pi, which will be pi divided by 2, multiplied by the integral of sine of you. Do you, and now we need to rewrite our boundaries in terms of of you. So what we will do is we will take both of our boundaries and plug them into our U substitution. Plugging in 0 will give us U is equal to 0 squared, and that is going to give us 0. And by plugging in the square root of pi divided by 6, you will equal to the square root of pi divided by 6 quantity squared, and that is going to simplify to pi divided by 6. Now what we are going to do is we are going to take the antiderivative of this of this integral. And the antiderivative of sine is going to be negative cosine. So we will be left with negative pi divided by 2. Multiplied by cosine of U, and this will be evaluated from 0 to pi divided by 6. And now what we're going to do is we're going to use the fundamental theorem of calculus part 2, to evaluate this antiderivative at both of the boundaries. So by plugging in the largest boundary, we will have negative pi divided by 2, multiplied by cosine of pi divided by 6, and then we will subtract negative pi divided by 2, multiplied by cosine of 0. And so now what we need to do is we need to simplify. Both of these quantities will simplify and factor the the pi divided by 2, and that will give us pi divided by 2 multiplied by the negative square root 3 divided by 2 plus 1. And if we were to factor out a 12 and reorder the terms, our final solution will be pi divided by 4, multiplied by the quantity, 2 minus the square root of 3. And this is going to be the volume for the region Q when it is rotated about the X-axis. And with that being said, the overall solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this type of 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. {Use of Tech} y = √sin^−1x,y = √π/2, and x=0; about the x-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.

{Use of Tech} y = √sin^−1x,y = √π/2, and x=0; about the x-axis