Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem (more here and the problem XIV.6:6 here on page 976)

Integrate $\int_A z dx dy dz$ in cylinder-coordinates when $$A=\{(x,y,z)\in\mathbb R^3 | x^2+y^2 \leq z \leq \sqrt{2-x^2-y^2}\}.$$

So $r^2=x^2+y^2$ and

$$\int_0^{2\pi} \int_a^b \int_{r^2}^{\sqrt{2-r^2}} rz dz dr d\rho$$ but what $a$ and $b$ are? I know for the max -case $0\leq z \leq \sqrt{2}$ and $0\leq r \leq 2^{0.25}$ if $z,r\in\mathbb R$. $$\int_0^{2\pi} \int_0^{2^{0.25}} \int_{r^2}^{\sqrt{2-r^2}} rz dz dr d\rho,$$ just this easy? With this specification, I got $\pi \left( \sqrt{2} - 0.5-\frac{1}{6} 2^{1.5} \right)$ as a solution. But if wrongly specified, it is not right. Perhaps, I need to find an expression for the $r$ in terms of the angle $\rho$ -- somehow from $x=r \cos(\rho)$, $y=r\sin(\rho)$ and $z=z$? I may be misunderstanding here the exercise, perhaps I am just over-engineering...

share|improve this question
    
With Joseki's notice, I got $\frac{7}{12} \pi$ as a solution (haven't verified). –  hhh Apr 30 '12 at 9:24

1 Answer 1

up vote 1 down vote accepted

I don't know what you mean by "the max-case". You need to integrate over the entire range of positive $r$ values for which the lower bound for $z$ is lower than the upper bound. The bounds are equal at $r^2=\sqrt{2-r^2}$, and thus $r=1$, so the integral with respect to $r$ should be over $[0,1]$.

share|improve this answer
    
Given $x^2+y^2 \leq z \leq \sqrt{2-x^2-y^2}$. In cyl-coords, $r^2 =x^2+y^2$ so $r^2 \leq z \leq \sqrt{2-r^2}$ so $r^4 \leq 2-r^2$ so $r^4+r^{2} -2\leq 0$ so $r^2=\pm 1$. And because $r\geq 0$ we have $0\leq r \leq 1$, yes you are right. I had a stupid mistake with this $r^2=\sqrt{2-r^2}$ (I for some reason thought that the $r$ was different on both sides but apparently not), now corrected thanks. –  hhh Apr 30 '12 at 9:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.