Setting up an integral wrong (solid of revolution).

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Mar 5th 2011, 04:05 PM
DannyMath

Setting up an integral wrong (solid of revolution).

Hi again,

I just used up 2 attempts on my assignment and it was wrong so I think I set up the integral wrong. Any help is appreciated:

The region bounded by y= 5 and y= x+(4/x)is rotated about the line x= -1 Find the volume of the resulting solid by any method.

I set up my integral as:

$2\pi\int_1^4 \! (x+1)(5-(x+(4/x))) \, \mathrm{d}x$

I tried to draw the graph and form the integral from it, is this the right direction?
Mar 5th 2011, 05:46 PM
Prove It

How did you get this integral?
Mar 5th 2011, 05:52 PM
Soroban

Hello, DannyMath!

Quote:

$\text{The region bounded by }y= 5\text{ and }y \:=\: x+\frac{4}{x}$
. . $\text{is rotated about the line }x= \text{-}1.$

$\text{Find the volume of the resulting solid by any method.}$

$\displaystyle \text{I set up my integral as: }\;2\pi\int_1^4 (x+1)\left(5-\left[x+\tfrac{4}{x}\right]\right)\,dx$

This is correct! . . . Good work!
Mar 5th 2011, 05:58 PM
TKHunny

No, you have it right. Be more careful in the evaluation of the integral.

The other way is much more difficult.

$\pi\int_{4}^{5}\left(\frac{y}{2}+\frac{\sqrt{y^{2}-16}}{2}+1\right)^{2}-\left(\frac{y}{2}-\frac{\sqrt{y^{2}-16}}{2}+1\right)^{2}\;dy$

I simply love to do it both ways - no matter the assignment. This gives greater practice and familiarity, in addition to verification.
Mar 5th 2011, 05:59 PM
DannyMath

Quote:

Originally Posted by Prove It

How did you get this integral?

Well I set it up visually from what my drawing looked like.

Quote:

Originally Posted by Soroban

Hello, DannyMath!

This is correct! . . . Good work!

Hmm, if this is indeed the correct integral, then maybe I'm calculating it wrong. My answer was -2pi(-9+[4ln(4)]) but it was marked wrong. I will attempt it again tomorrow morning to see if I get a different answer. But I am glad that the integral was correct as this is the hardest part of the problem :P
Mar 5th 2011, 06:06 PM
DannyMath

Quote:

Originally Posted by TKHunny

No, you have it right. Be more careful in the evaluation of the integral.

The other way is much more difficult.

$\pi\int_{4}^{5}\left(\frac{y}{2}+\frac{\sqrt{y^{2}-16}}{2}+1\right)^{2}-\left(\frac{y}{2}-\frac{\sqrt{y^{2}-16}}{2}+1\right)^{2}\;dy$

I simply love to do it both ways - no matter the assignment. This gives greater practice and familiarity, in addition to verification.

Wow that looks hairier than my grandpa's nose, but I agree about being familiar with as many methods as possible. As myself always says, knowledge begets knowledge!
Mar 5th 2011, 06:41 PM
TKHunny

Quote:

Originally Posted by DannyMath

-2pi(-9+[4ln(4)])

Yup, you're a little short, there. Should be $8\cdot\pi\cdot(3-ln(4))$