Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.

y= 2e^(−x), y= 2, x= 6; about y = 4.

How exactly do you set up the integral? I know that I am supposed to use the washer method but I can't figure out the limits of integration or what I use for f(x) and g(x).

2e^-x has a y-intercept of 2, so you have a sort of triangular region with vertices at

(0,2),(6,2),(6,2e^-6)

using washers, each washer has an area

π(R^2-r^2)
where r = 4-2 = 2
and R = 4-y = 4-2e^-x

Now the integral is just

π∫[0,6] (4-2e^-x)^2-2^2) dx

You can also use shells, where each shell of thickness dy has a height 6-x = 6-ln(2/y) = 6+ln(y/2)

and the volume is thus

∫[2e^-6,2] 2πrh dy
where r = 4-y and h = 6+ln(y/2)

That's a bit trickier, using integration by parts, but should come out the same.

Thank you. The answer would be pi(58+16e^-6-2e^-12)correct?

That's what I get.

For shells,

∫[2e^-6,2] 2π(4-y)(6+log(y/2)) dy

comes out the same. Amazing!

To set up the integral using the washer method, we consider that the solid obtained by rotating the region bounded by the given curves about the line y = 4 consists of infinitesimally thin washers.

First, let's find the limits of integration. The region bounded by the curves y = 2e^(-x), y = 2, and x = 6 is a finite region between x = 0 and x = 6. Therefore, the limits of integration for x will be from 0 to 6.

Next, let's determine the functions f(x) and g(x) that represent the upper and lower boundaries of the region being revolved.

Since the curve y = 2e^(-x) is the upper boundary and the line y = 2 is the lower boundary, we can use these equations to obtain the functions f(x) and g(x), respectively.

f(x) = 2e^(-x)
g(x) = 2

Now, we can visualize the situation considering each infinitesimally thin washer:

For each value of x, we have an infinitesimally thin washer with an outer radius equal to the distance from the line y = 4 to the curve y = 2e^(-x) (f(x)), and an inner radius equal to the distance from the line y = 4 to the curve y = 2 (g(x)).

The volume V of each washer can be calculated as V = π(R^2 - r^2)h, where R is the outer radius, r is the inner radius, and h is the height of the washer.

In this case, the outer radius R is 4 - f(x) and the inner radius r is 4 - g(x). Since we are revolving around the line y = 4, the height h of each washer is dx (an infinitesimal change in x).

Therefore, the volume dV of each washer is dV = π((4 - f(x))^2 - (4 - g(x))^2)dx.

To find the total volume V, we integrate the volumes of all the washers from x = 0 to x = 6:

V = ∫[0,6] π((4 - f(x))^2 - (4 - g(x))^2)dx

Substituting the expressions for f(x) and g(x), we have:

V = ∫[0,6] π((4 - 2e^(-x))^2 - (4 - 2)^2)dx

Now, you can evaluate this integral to find the volume V of the solid obtained by rotating the region about the line y = 4.