Question

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).

Answer 1

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.

Answer 2

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

Answer 3

That's what I get.

For shells,

∫[2e^-6,2] 2π(4-y)(6+log(y/2)) dy

comes out the same. Amazing!