Exercises 83–86 are about the infinite region in the first qua...

Question

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.

86. Find the volume of the solid generated by revolving the region about the x-axis.

Accepted Answer

Hello. In this video we are going to consider the region r in the first quadrant bounded above by the curve y is equal to e to the power of -2 x below by the x axis, and between the boundaries of x is equal to 0 and x is equal to infinity. We want to find the volume of the solid generated when r is revolved about the x axis. Now, since we're evolving this region about the X axis, and we're integrating with respect to X, we can approach this problem by finding the volume using the method of the disc method. So let's go ahead and define our volume. Now the volume for the disc method is defined as the definite integral from A to B of pi multiplied by a radius squared d x. Now, in this case, our boundaries are going to be from 0 to infinity, since we're trying to find an infinite area. So we're integrating from 0 to infinity. Now we need to define the radius. Now in this case here, the radius is going to be defined as the curve, and radius is going to be e to the power of -2 x. Now, if we were to square both sides of this radius, the radius squared is going to be e to the power of -4 x. So, if we were to apply this information to the volume equation, our volume is going to be defined as the improper integral from zero to infinity of pi multiplied by e to the power of -4 XDX. So let's go ahead and evaluate this integral by using the method of improper integrals. Now the first thing we want to do is we want to go ahead and create a substitution for an infinite boundary. We will sit, we will make a substitution with a variable, let's say B. So we'll take the limit as b approaches infinity. Of the infinite integral 0 to b of pi multiplied by e to the power of -4 x d x. Now if we were to take or if we were to integrate at this point, the antiderivative of this integral will be the limit as b approaches infinity of negative pi divided by 4 e to the power of -4 x, and this is going to be evaluated from 0 to b. Now if we were to evaluate this at the boundaries, we will have the limit as b approaches infinity of negative pi divided by 4 e to the power of -4 x minus, oh sorry, not 4 x, it should be 4b, apologies. So E to the power of -4B minus pi divided by 4, E-4 multiplied by 0. Now this is going to simplify to leave us with negative pi divided by 4 e to the power of negative infinity. Plus pi divided by 4. Now, this first term is going to simplify to just be zero, and so what we're left with is 0 plus pi divided by 4, which gives us a total volume under the curve of pi divided by 4. This is going to be the solution to the problem, and the overall solution is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.86. Find the volume of the solid generated by revolving the region about the x-axis.

Watch next

Exercises 83–86 are about the infinite region in the first quadrant between the curve y = e^(-x) and the x-axis.
86. Find the volume of the solid generated by revolving the region about the x-axis.