42-47. Volumes of Solids Find the volume of the solid generate...

Question

42-47. Volumes of Solids Find the volume of the solid generated when the given region is revolved as described.

44. The region bounded by f(x) = sin(x) and the x-axis on [0, π] is revolved about the y-axis.

Accepted Answer

Hello. In this video, we are going to be finding the volume of the solid generated when the region bounded by F of X equal to cosine of X, and the X-axis on the interval from 0 to pi divided by 2 is revolved about the y axis. Now, in order to approach this problem, the first thing we should do is we should go ahead and graph the given information. Now, we are going to graph cosine of X, but we are only graphing the portion on the interval from 0 to pi divided by 2. Now, for the cosine function, cosine starts at 1 on the Y axis at the value X is equal to 0, and at pi divided by 2, cosine has the value of 0. So the shape of this curve is going to look like this. So this curve here is going to be defined as Y is equal to cosine of X. Now, this curve here. And the x axis is bounding. This region between 0 and pi divided by 2. Furthermore, we are told that we are going to revolve this region about the y axis. Now, in order to approach this problem, and in order to define the volume, if we were to draw a quick sketch of the reflected region, the reflected region will look roughly like this. Now if we try to use the disk and washer method, we are going to have a little overlap between the distances between the two sides because the distance between the left and right hand curve are going to be defined as the same function. This is a characteristic that tells us that we should approach this problem by using the shell method. So, we are going to approach and find the volume by using the shell method. By the shell method, our volume is going to be defined as two pi multiplied by an integral from A to B of the radius of the shell multiplied by its height, and in this case here because we are revolving about the y axis, we are going to integrate with respect to X. Now, the first thing we're going to do is we're going to define the bounds of integration. But in this case here, the bounds of integration were given to us in the problem. The bounds of integration are going to be from 0 to pi divided by 2. Now, what about the radius of the shell? Now, the radius of the shell is just going to be some measure of X along the X axis. So radius in this case, is going to be defined as X. Now, the height of the shell is going to be defined as the distance from the X-axis to the uppermost curve, and that distance is going to be defined as cosine of X. So height in this case is going to be cosine of X. And so now that we have the 3 pieces we need, we are going to construct the volume integral. So the volume for this region is going to be defined as 2 pi multiplied by the integral from 0 to pi divided by 2 of our radius X multiplied by our height, cosine of X, and this will be integrated with respect to X. Now, all we need to do in this case is we need to go ahead and solve the following integral in order to get the volume of the region. Now, because we have X multiplied by cosine of X, we will solve this integral by using integration by parts. We are going to allow U to equal to X, and by taking the derivative of both sides of this equation, we will have DU is equal to DX. We will define our DV function to be cosine of X, and by integrating both sides of the DV function, we will have V is equal to sine of X. And so by putting everything together by the integration by parts, we will have 2 pi multiplied by the quantity U times V, which is sin or X multiplied by sine of X, evaluated from 0 to pi divided by 2. Minus the integral from 0 to pi divided by 2 of VDU which is going to be sin of XDX. Now we can go ahead and take the antiderivative of sin of X, that is going to be negative cosine of X. So we are left with 2 pi multiplied by the quantity, X, sin of X. Plus cosine of X, and this will be evaluated from 0 to pi divided by 2. And so now what we are going to do is we are going to use the fundamental theorem of calculus part 2, in order to solve the remaining of the problem. So we're going to plug in the values of both of the boundaries. That is going to leave us with 2 pi multiplied by the quantities of pi divided by 2 plugged into the antiderivative, that is going to leave us with pi divided by 2, multiplied by sine of pi divided by 2, plus cosine of pi divided by 2. Minus the quantity. Of 0 multiplied by sin of 0, plus cosine of 0. Now, there's a bit of simplification we are going to do here. This term is just going to simplify to zero, because anything multiplied by 0 is 0. Cosine has a value of 1 at the angle of zero. Cosine has a value of 0 at the angle pi divided by 2, and sine has a value of 1 at the angle pi divided by 2. And so by applying these simplifications, we are left with 2 pi multiplied by the quantity, pi divided by 2 minus 1. And finally, if we factor out a 12 from the quantity, that will leave us with pi multiplied by pi minus 2, and this is going to be the volume of the region when it is revolved about the y axis. 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 find this volume, and we will go ahead and see you all in the next video.

42-47. Volumes of Solids Find the volume of the solid generated when the given region is revolved as described.44. The region bounded by f(x) = sin(x) and the x-axis on [0, π] is revolved about the y-axis.

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42-47. Volumes of Solids Find the volume of the solid generated when the given region is revolved as described.
44. The region bounded by f(x) = sin(x) and the x-axis on [0, π] is revolved about the y-axis.