# Homework Help: Can't get this one

1. Oct 10, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

Find the exact volume of the region bounded by:

y = e^3x
y = e^x
x=0
x=1

Revolved around x axis

2. Relevant equations

3. The attempt at a solution

The partitions will be disks of radius e^3x - e^x and height of delta x.

Therefore, the volume of a partition is

pi(e^3x - e^x)^2 delta x

Integrating this from 0 to 1 gives me a final integral of:

$\pi \int^{1}_{0} e^{6x} - e^{2x} dx$

Which evaluated for exact gives me:

$\pi(\frac{e^{6} - 1}{6} - \frac{e^{2} + 1}{2})$

But I am incorrect. Why?

2. Oct 10, 2011

### SammyS

Staff Emeritus

As it turns out, that's what you have in your integral expression. (But two wrongs don't make a right.)

Check your signs when you substitute the limits of integration into your anti-derivative.

3. Oct 10, 2011

### HallsofIvy

No, they won't. In fact there are NO "disks". The region between the two graphs form a "washer". One way to do this is to find the volume wnem the disk formed by $e^{3x}$ is rotated around the x-axis, find the volume when the disk formed by $e^x$ is rotated around the x-axis, and then subtracting

No. $\pi\int e^{3x}dx- \pi\int e^x dx= \pi \int (e^3x- e^x) dx$