Finding the volume of solids..

Hi this is regarding volume of solids, it's the basics but I can't get the answers and it's frustrating me! Any help will be very much appreciated, really!

1.) Find the volume of solid formed when y= $\displaystyle x+\frac{1}{x}$ is rotated for $\displaystyle 1\leq x \leq3$

I integrated $\displaystyle (x+\frac{1}{x})^{2}$ from 1 to 3 and got 12pi units^3 as the answer, my answer after integration was pi($\displaystyle \frac{x^3}{3}+2x+\frac{1}{x})$..I can't see any possible errors in my working, I must have integrated wrongly..

2.) Find the volume of solid formed when y= $\displaystyle \frac{x^3}{x^{2}+1}$ is rotated for xE[1,3]

I integrated $\displaystyle \frac{x^6}{x^{4}+1}$ from 1 to 3.

I chose to use trig substitution, so I got $\displaystyle x=\tan \theta$

I had changed $\displaystyle x^{4}+1$ to $\displaystyle \sec^4 \theta$ since I thought $\displaystyle \tan^2 \theta + 1 = \sec^2 \theta$, then $\displaystyle \tan^4 \theta + 1 = \sec^4 \theta$..

Simplifying, I ended up with $\displaystyle \pi\tan^2 \theta\sin^4 \theta$

Also, I thought since $\displaystyle \sin^2 \theta + \cos^2 \theta=1$, then $\displaystyle \sin^4 \theta + \cos^4 \theta=1$

so I got $\displaystyle \pi\tan^2 \theta\(1-cos^4 \theta)$..

Separating into two integrals and changing tan to sec again, I ended up with

$\displaystyle \int^3_1\frac{3pi}{2}-\frac{cos\2\theta}{2}\dx $..

Integrating I got

$\displaystyle \frac{1}{2}\pi\+\(3\theta+\frac{1}{2}\sin\2\theta)$, from 1 to 3..

Are my steps correct so far?

3.) A hemispherical bowl of radius 8cm is filled to a depth of 3cm. Find the volume of water in the bowl. The equation is $\displaystyle \x^2+y^2=64$

I integrated from 0 to 3, the equation $\displaystyle 64-y^2$..

Please assist me I really am at a loss as to why my answers are all wrong..

Thank you very much!

Re: Finding the volume of solids..

(1) I assume the curve is rotated about the x-axis ...

$\displaystyle V = \pi \int_1^3 \left(x + \frac{1}{x}\right)^2 \, dx$

$\displaystyle V = \pi \int_1^3 x^2 + 2 + \frac{1}{x^2} \, dx$

$\displaystyle V = \pi \left[ \frac{x^3}{3} + 2x {\color{red}- \frac{1}{x}} \right]_1^3$

Re: Finding the volume of solids..

(2) note ...

$\displaystyle \left(\frac{x^3}{x^2+1}\right)^2 \ne \frac{x^6}{x^4+1}$

Re: Finding the volume of solids..

For 2 is it x^6/(x^2+1)^2? I integrated with trig sub and got tan2x-x...please show me how you'd integrate, ibe a feeling im wrong..

Re: Finding the volume of solids..

long division ...

$\displaystyle \frac{x^6}{x^4+2x^2+1} =$

$\displaystyle x^2 - 2 + \frac{3x^2+2}{(x^2+1)^2} =$

$\displaystyle x^2 - 2 + \frac{3x^2+3 - 1}{(x^2+1)^2} =$

$\displaystyle x^2 - 2 + \frac{3(x^2+1)}{(x^2+1)^2} - \frac{1}{(x^2+1)^2} =$

$\displaystyle x^2 - 2 + \frac{3}{x^2+1} - \frac{1}{(x^2+1)^2}$

integrating the first three terms is rather straightforward ... I would use the trig substitution $\displaystyle x = \tan{t}$ for the last term.

Re: Finding the volume of solids..

Thank you!

I wonder if there is something wrong with my method of long division, I got a different answer from yours..

$\displaystyle \frac{x^6}{(x^2+1)^3} = x^2 - \frac{2x^2-1}{x^4+2x^2+1}$

I used partial fractions to tackle the fraction..

$\displaystyle \frac{x^6}{(x^2+1)^3} = x^2 - \frac{2}{(x^2+1)}-\frac{3}{(x^2+1)^2}$

Thank you so much, really appreciate your help!

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Re: Finding the volume of solids..

Quote:

Originally Posted by

**Tutu** I wonder if there is something wrong with my method of long division ...

looks that way ...