# Thread: Finding the volume of solids..

1. ## 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= $x+\frac{1}{x}$ is rotated for $1\leq x \leq3$
I integrated $(x+\frac{1}{x})^{2}$ from 1 to 3 and got 12pi units^3 as the answer, my answer after integration was pi( $\frac{x^3}{3}+2x+\frac{1}{x})$..I can't see any possible errors in my working, I must have integrated wrongly..

2.) Find the volume of solid formed when y= $\frac{x^3}{x^{2}+1}$ is rotated for xE[1,3]
I integrated $\frac{x^6}{x^{4}+1}$ from 1 to 3.
I chose to use trig substitution, so I got $x=\tan \theta$
I had changed $x^{4}+1$ to $\sec^4 \theta$ since I thought $\tan^2 \theta + 1 = \sec^2 \theta$, then $\tan^4 \theta + 1 = \sec^4 \theta$..
Simplifying, I ended up with $\pi\tan^2 \theta\sin^4 \theta$
Also, I thought since $\sin^2 \theta + \cos^2 \theta=1$, then $\sin^4 \theta + \cos^4 \theta=1$
so I got $\pi\tan^2 \theta\(1-cos^4 \theta)$..
Separating into two integrals and changing tan to sec again, I ended up with
$\int^3_1\frac{3pi}{2}-\frac{cos\2\theta}{2}\dx$..
Integrating I got
$\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 $\x^2+y^2=64$
I integrated from 0 to 3, the equation $64-y^2$..

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

Thank you very much!

2. ## Re: Finding the volume of solids..

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

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

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

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

3. ## Re: Finding the volume of solids..

(2) note ...

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

4. ## 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..

5. ## Re: Finding the volume of solids..

long division ...

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

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

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

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

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

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

6. ## 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..

$\frac{x^6}{(x^2+1)^3} = x^2 - \frac{2x^2-1}{x^4+2x^2+1}$
I used partial fractions to tackle the fraction..

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

Thank you so much, really appreciate your help!

7. ## Re: Finding the volume of solids..

Originally Posted by Tutu
I wonder if there is something wrong with my method of long division ...
looks that way ...