Why Is My Calculation of the Volume Using the Shell Method Incorrect?

[Shakenbake158]

Homework Statement

Hey guys, I am working an integration problem and am having a hard time.
This is the problem : Let R be the region bounded by Y = 2x^(-3/2), Y = 2, Y = 16, and X = 0. Use the shell method to find the volume of the solid generated when R is revolved about the X-Axis.

The radius is just Y since it's about an axis.
The bounds are from 2 to 16.
I don't know if I am solving for X wrong, when I solve for X, I get X = (1/2)Y^(-2/3)
So wouldn't my integral be 2∏∫(Y)((1/2)Y^(-2/3)dy from 2 to 16?
the anser is 90∏, and that is not what I am getting :/

Homework Equations

So I know the shell method is 2∏∫(radius)(height).
Since its about the X-Axis, we are integrating with respect to Y, which means we have to solve for X.

The Attempt at a Solution

The radius is just Y since it's about an axis.
The bounds are from 2 to 16.
I don't know if I am solving for X wrong, when I solve for X, I get X = (1/2)Y^(-2/3)
So wouldn't my integral be 2∏∫(Y)((1/2)Y^(-2/3) from 2 to 16?
the anser is 90pi, and that is not what I am getting :/

 

[SammyS]

Homework Statement

Hey guys, I am working an integration problem and am having a hard time.
This is the problem : Let R be the region bounded by Y = 2x^(-3/2), Y = 2, Y = 16, and X = 0. Use the shell method to find the volume of the solid generated when R is revolved about the X-Axis.

The radius is just Y since it's about an axis.
The bounds are from 2 to 16.
I don't know if I am solving for X wrong, when I solve for X, I get X = (1/2)Y^(-2/3)
So wouldn't my integral be 2p times the integral of (Y)((1/2)Y^(-2/3) from 2 to 16?
the anser is 90pi, and that is not what I am getting :/

Homework Equations

So I know the shell method is the integral of 2pi(radius)(height).
Since its about the X-Axis, we are integrating with respect to Y, which means we have to solve for X.

The Attempt at a Solution

The radius is just Y since it's about an axis.
The bounds are from 2 to 16.
I don't know if I am solving for X wrong, when I solve for X, I get X = (1/2)Y^(-2/3)
So wouldn't my integral be 2pi times the integral of (Y)((1/2)Y^(-2/3) from 2 to 16?
the anser is 90pi, and that is not what I am getting :/

Hello Shakenbake158. Welcome to PF.

You're either solving for x incorrectly, or you're incorrectly expressing the result.

(y/2) should be raised to the -2/3 power, not y alone.

 

[Shakenbake158]

Hello Shakenbake158. Welcome to PF.

You're either solving for x incorrectly, or you're incorrectly expressing the result.

(y/2) should be raised to the -2/3 power, not y alone.

Thank you for the greeting!

I thought that I might be solving for X wrong.

So now I have 2∏∫(y)((y/2)^(-2/3))dy

My algerbra is failing me, how do I multiple these two terms?

 

[SammyS]

\displaystyle \left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}


\displaystyle a^m a^n=\left(a\right)^{(m+n)}

 

[Shakenbake158]

\displaystyle \left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}


\displaystyle a^m a^n=\left(a\right)^{(m+n)}

Thank you so much for the help. I was finally able to get 90∏.

I was just making algebraic mistakes.

I appreciate the help.