I'm having the following integral, and I'm being asked to convert the integration boundaries to cylindrical coordinates.

Original Integral Boundaries

I've figured out that on XY-plane it's an ellipse having the following equation:

$$ \frac{(x-\frac{a}{2})^2}{\frac{9a^2}{4}}+\frac{y^2}{\frac{9a^2}{2}} = 1 $$

I realize that $ 0\leq \theta \leq \frac{\pi}{2}$ and $0 \leq z \leq r^2(\theta)$

But I don't seem to be able to reach an expression for $r(\theta)$.

Answer says that $r(\theta) = 2acos(\theta)$.

Any ideas? Thank you.


1 Answer 1


Lets analyze the bounds in the integral: \begin{align} 0 &\le x \le 2a\\ 0 &\le y \le \sqrt{2ax-x^2}\\ 0 &\le z \le x^2+y^2\\ \end{align} By squaring the second equation, we have $$ 0\le y^2 \le 2ax-x^2\quad \Rightarrow\quad 0\le (x-a)^2+y^2 \le a^2, $$ which represents a disc of radius $a$, centered at $(a,0)$. This gives us a more clear image of what the bounds on $x$ and $y$ represent (draw the disc and read the bounds): it is the part of this disc in the $y\ge 0$ plane.

The disc $(x-a)^2+y^2 \le a^2$ has polar equation $2a\cos\theta$ (just replace $x$ and $y$ by $r\cos\theta$ and $r\sin\theta$, respectively.

With this information, it is straightforward to express the bounds in cylindrical coordinates: \begin{align} 0 &\le \theta \le \frac{\pi}{2}\\ 0 &\le r \le 2a\cos\theta\\ 0 &\le z \le r^2\\ \end{align}

  • $\begingroup$ Great point. I've squared the two inequalities, and summed them and therefore I received an ellipse and a more difficult expression than you did. Thank you. $\endgroup$
    – Taru
    Commented Jun 29, 2016 at 15:51

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