# [SOLVED] Volume of Solid Of Revolution....y=sqrt(x) and y=x, around x=5

• May 28th 2009, 10:31 AM
calc101
[SOLVED] Volume of Solid Of Revolution....y=sqrt(x) and y=x, around x=5
Hello,

I have been trying to solve the following problem:

Consider the solid obtained by rotating the region bounded by the given curves about the line x = 5. The curves are: $y=\sqrt{x}$ and $y=x$. Find the volume V of this solid.

Facts: The only points of interesction are at (0,0) and (1,1).

Attempted Solution:

We need to use the washer method. However, since the region is being rotated around the line $x=5$, the radius of each circle will be $r = r + 5$. We also need to integrate from 0 to 1 along y. Therefore, my solution looked something like this:

$V = pi \int (y^2+5)^2dy - pi \int (y+5)^2dy$

expanding....

$V = pi \int y^4+10y^2+25dy - pi \int y^2+10y+25dy$

simplify...

$V= pi \int y^4+9y^2-10y dy$

integrate...

$V = pi(\frac{y^5}{5} + 3y^3 - 5y^2)$

$V = pi(\frac{1}{5} + 1 - \frac{5}{2})$

$V = \frac{-21}{10}pi$

The problem with my solution is that volume can not be negative. Can some one please explain what is wrong with my solution?

Thanks,
• May 28th 2009, 12:17 PM
Soroban
Hello, calc101!

Did you make a sketch?

Quote:

Consider the solid obtained by rotating the region bounded by the given curves
about the line $x = 5.$ .The curves are: $y\:=\:\sqrt{x}\text{ and }y\:=\:x$.
Find the volume $V$ of this solid.

Facts: The only points of interesction are at (0,0) and (1,1).

Attempted Solution:

We need to use the washer method.
However, since the region is being rotated around the line $x=5$,
the radius of each circle will be $r = r + 5$. . . . . no
We also need to integrate from 0 to 1 along y.

Code:

      |                :       |                *       |      ...*      :       |  *:::/        :       | *:::/          :       |*::/            :       |:/              :     - * - - - - - - - - + - -       |                5

Our functions are: . $x \,=\,y^2,\;x \,=\,y$

The outer radius is: . $r_1 \:=\:5-y^2$
The inner radius is: . $r_2 \:=\:5-y$

Hence: . $V \;=\;\pi\int^1_0(5-y^2)^2\,dy - \pi\int^1_0(5-y)^2\,dy$

• May 28th 2009, 12:35 PM
Danneedshelp
You integrated wrong. Your set up needs to be changed anyways...

Try....
$
V \;=\;\pi\int^1_0y^4-11y^2+10y\,dy
$

Oh man, didn't see the guy above me already worked it out. His work is the same as mine though. You just messed up the radius calculation...
• May 28th 2009, 07:21 PM
calc101
gosh, i feel stupid....why of course....the radii is five points from the origin minus the change produced by the functions...

I did sketch it, but it didn't quite click...

thanks a lot guys...