Let R be the region bounded by the following curves. Find the ...

Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=2x,y=0 , and x=3; about the x-axis (Verify that your answer agrees with the volume formula for a cone.)

Graph showing region bounded by y=2x, y=0, and x=3, shaded area labeled R, with axes and point (3,6).

Accepted Answer

Hello. In this video, we are going to let T be the region bounded by Y is equal to X, Y is equal to 0, and X is equal to 2. We want to find the region or the volume of the solid generated when T is revolved about the X axis. Now, if we take a look at the given shape, when we revolve the shape about the X-axis, we are going to end up with a symmetric shape that has no gaps within it. And so since there are no gaps when revolving about the X-axis, we can go ahead and approach this problem by using the disk method. So, by using the disk method, the volume is going to be defined as pi multiplied by a definite integral from A to B, of a radius squared, and in this case here, because we are revolving about the X axis, we are going to integrate with respect to X. And so in order to find the volume, what we need to do is to to define our bounds of integration and to define the radius of the disk. Now, for the bounds of integration, we want to go ahead and find the smallest and largest value of X in the region. Now, notice that the regions intersect each other at the origin. What this means is that the smallest value of X that we are going to use for our boundary is going to be X is equal to 0. The largest value of X is going to be defined as the constant function X is equal to 2. So that means that we're going to be integrating from 0 to 2. Now we need to define the radius for the integral or for the disk method. Now the radius is going to be the distance from the X axis to the uppermost curve of the region, and that is going to be defined as Y is equal to X. So our region, our radius is going to be X. And so now that we have our boundaries and our radius, we can go ahead and construct the integral for the volume. The volume is going to be defined as pi multiplied by the integral from 0 to 2 of our radius squared, which is X2 DX. Now what we are going to do is we are going to solve for the integral in order to find the volume. Now, we can go ahead and take the antiderivative of this integral right away. By taking the antiderivative, we'll be left with pi multiplied by the quantity, 1/3 X cubed, and this will be evaluated from 0 to 2. And by combining the constant terms together, we will have pi divided by 3, multiplied by X cubed, evaluated from 0 to 2. Now, what we are going to do is we are going to use the fundamental theorem of calculus part 2 to further evaluate the volume. So by plugging in the values of the bounds, we will have 3 multiplied by 2 cubed minus 3 multiplied by 0 cubed. Now, when we plug in 0, we just get the value of 0, and 2 of 3 evaluates to be 8. And 8 multiplied by pi divided by 3, is going to give us 8 pi divided by 3, and this is going to be the volume of the region when it is revolved about the X 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.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.y=2x,y=0 , and x=3; about the x-axis (Verify that your answer agrees with the volume formula for a cone.)

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Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.
y=2x,y=0 , and x=3; about the x-axis (Verify that your answer agrees with the volume formula for a cone.)