Double integral on cone surface with cylindrical coordinates

[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

 

[member 553083]

!For this integral, you can use the substitution u = r^2. This will give us: \sqrt{2} \int_{0}^{2\pi} \int_{1}^{4}(u^2 + u \cos(\theta)^2 \sin(\theta)^2) \: du \: d\thetaIntegrating with respect to u first will give us: \sqrt{2} \int_{0}^{2\pi} \left[\frac{u^3}{3} + \frac{u^2 \cos(\theta)^2 \sin(\theta)^2}{2}\right]_1^4 \: d\thetaEvaluating this integral and multiplying by \sqrt{2} will give us the final answer: \sqrt{2} \left[\frac{64}{3} + 8\sin(\theta)^2 \cos(\theta)^2\right]_0^{2\pi}The final answer is therefore: \sqrt{2} \left(\frac{64\pi}{3} + 16\pi\right)