Surface area and volume Let f(x) = 1/3 x³ and let R be the reg...

Question

Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].

c. Find the volume of the solid generated when R is revolved about the x-axis.

Accepted Answer

Hello. In this video, we are going to let the region bounded by F of X equal to X the 4th power and the X-axis on the interval from 0 to 1 be revolved about the X axis. What is the volume of the resulting solid? Now, in order to approach this problem, let's go ahead and first graph what is given to us. We know that we are working on the interval from 0 to 1, so that is going to be the right half of the axis, both quadrants 1 and 4. So, we will go ahead and bound 1 and 0 on the number line. Now we're going to draw the gra the graphs that are bounding this region. First, we know that the region is bounded by the X axis. So let's go ahead and highlight that. And we are going to let the region be bounded by X 4th power. Now, X the 4th power is a cubic function. That stretches from 0 to 1. So this is going to be the shape of F of X equal to X to the 4th power. So, the region that we are revolving is going to be the shaded region here. Now, one more thing that we are told is that we are revolving the shape about the X axis. So, the other half of the shape is going to look similar to this. Now, notice that if we were to reflect a shape about the X-axis, then we get a solid shape. What this means is that in order to find the volume, we can approach this by using the disk method. Now, for the volume, the disk, by the disc method, the volume is defined as the following. Volume is equal to pi, multiplied by the definite integral from A to B of the radius of the disk squared. And in this case here, because we're rotating about the X axis, we are going to integrate with respect to X. So what we need to do is we need to go ahead and define our radius. Now radius in this case is going to be the distance from the X axis to the uppermost curve, and that's just going to be X to the 4th power. And we are given the bounds of integration in the problem, we are integrating from 0 to 1. So our volume integral is going to be defined as pi, multiplied by the integral from 0 to 1 of X to the 4th power squared DX. All we need to do now is solve the equation to get the volume of the shape. By squaring the integrand, we will be left with pi multiplied by the integral from 0 to 1 of X 8 DX. Then, by taking the antiderivative, we will be left with pi multiplied by 1/9, multiplied by X 9th power, and this will be evaluated from 0 to 1. Furthermore, we can rewrite this as pi divided by 9 multiplied by x to the 9th power. Evaluated from 0 to 1. And by using the fundamental theorem of calculus part 2, if we were to plug in the upper and lower bound, we will be left with pi divided by 9 minus 0, which is going to simplify to be pi divided by 9. This is going to be the volume of the shape 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 approach this problem, and we will go ahead and see you all in the next video.

Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].c. Find the volume of the solid generated when R is revolved about the x-axis.

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Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].

c. Find the volume of the solid generated when R is revolved about the x-axis.