Double Integration: Evaluating z^2 = x^2 + y^2

[rolylane]

Hi
I've been working on a problem and I'm nearly there but I'm struggling with the integration part at the end and was hoping you might be able to help if you have the time. The original question was

$$\int \int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS$$
Evaluated on the region of $$z^2 = x^2 + y^2$$ between z=1 and z=2.
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Then by substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I got $$\sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$$

and then I'm stuck. Any help or advice would really be appreciated
Thanks

 

[NateTG]

This can be cracked with trig identities, or done by parts.

 

[unplebeian]

I haven't done the steps but are you sure the integrand contains cos (theta^2) and sin (theta^2) or is it cos^2 (theta) (cos squared theta) ?

 

[blochwave]

Same thing, he didn't write cos(theta^2), they're outside the parentheses

 

[CompuChip]

I haven't done the steps but are you sure the integrand contains cos (theta^2) and sin (theta^2) or is it cos^2 (theta) (cos srquared theta) ?

It says $\cos(\theta)^2$ which I read as $\cos^2(\theta)$ as opposed to $\cos(\theta^2)$.

You might replace $\cos^2\theta = 1 - \sin^2\theta$, then two of the integrals become very easy, and one of them becomes $\sin^4\theta$ for which there are tricks.

 

[unplebeian]

Same thing, he didn't write cos(theta^2), they're outside the parentheses

Well, then just go brute force and then apply by parts. I got int( sin^2(2 x theta))

After simplyfying and using by parts I get int( 2*theta*sin(4theta))

By parts again.

 

[NateTG]

It says $\cos(\theta)^2$ which I read as $\cos^2(\theta)$ as opposed to $\cos(\theta^2)$.

You might replace $\cos^2\theta = 1 - \sin^2\theta$, then two of the integrals become very easy, and one of them becomes $\sin^4\theta$ for which there are tricks.

Actually, I was thinking...
$$\cos \theta \sin \theta = \frac{1}{2} \sin \left( 2 \theta \right)$$