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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:
$\displaystyle 2\pi\int_1^4 \! (x+1)(5-(x+(4/x))) \, \mathrm{d}x$
I tried to draw the graph and form the integral from it, is this the right direction?
How did you get this integral?
Hello, DannyMath!
Quote:
$\displaystyle \text{The region bounded by }y= 5\text{ and }y \:=\: x+\frac{4}{x}$
. . $\displaystyle \text{is rotated about the line }x= \text{-}1.$
$\displaystyle \text{Find the volume of the resulting solid by any method.}$
$\displaystyle \displaystyle \text{I set up my integral as: }\;2\pi\int_1^4 (x+1)\left(5-\left[x+\tfrac{4}{x}\right]\right)\,dx$
This is correct! . . . Good work!
No, you have it right. Be more careful in the evaluation of the integral.
The other way is much more difficult.
$\displaystyle \pi\int_{4}^{5}\left(\frac{y}{2}+\frac{\sqrt{y^{2}-16}}{2}+1\right)^{2}-\left(\frac{y}{2}-\frac{\sqrt{y^{2}-16}}{2}+1\right)^{2}\;dy$
I simply love to do it both ways - no matter the assignment. This gives greater practice and familiarity, in addition to verification.
Well I set it up visually from what my drawing looked like.Quote:
Originally Posted by Prove It
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 :PQuote:
Originally Posted by Soroban
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!Quote:
Originally Posted by TKHunny
Yup, you're a little short, there. Should be $\displaystyle 8\cdot\pi\cdot(3-ln(4))$Quote:
Originally Posted by DannyMath
