Find Volume of Region Bounded by y=e^3x & y=e^x

[1MileCrash]

Homework Statement

Find the exact volume of the region bounded by:

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

Revolved around x axis

Homework Equations

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?

 

[SammyS]

They're (vertical) washers, with outer radius e^3x and inner radius e^x.

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.

 

[HallsofIvy]

Homework Statement

Find the exact volume of the region bounded by:

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

Revolved around x axis

Homework Equations

The Attempt at a Solution

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

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

Therefore, the volume of a partition is

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

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

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?