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!