Question

Integral in cylindrical coordinates

Answer 1

Please do not comment unless you are making a serious attempt to answer the question. I don't go around leaving useless spam comments in other people's posts so please extend that level of courtesy to other people.

Answer 2

The inner most integral is with respect to theta, so you'll be integrating cos^2(theta) which on the surface looks impossible. But if you use the trig identity
\(\Large \cos^2(\theta) = \frac{1}{2}\left(1+\cos(2\theta)\right)\)
then it's doable. You'll use u-substitution letting u = 2theta

The z^2 and r^2 terms are constants when theta is the variable you're integrating with respect to. So that means the z^2 at the end turns into theta*z^2, and you'll just pull out the coefficient r^2.

Here's a list of trig identities
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
The trig identity I used is at the top right corner of page 2

Answer 3

Doing it in order presented:

\(I= \int\limits_{z = 0}^4 \int\limits_{r = 0}^{\sqrt z} \int\limits_{\theta = 0}^{2 \pi} r^3 cos^2 \theta + r z^2 ~ ~ d\theta ~dr ~dz \)

From Pythagorean Identity and symmetry:

\(\int\limits _0^{2 \pi} cos^2 \theta + sin^2 \theta ~ d \theta = 2 \pi \qquad \int\limits _0^{2 \pi} cos^2 \theta ~ d \theta= \int\limits _0^{2 \pi} sin^2 \theta ~ d \theta\)

\(\implies \int\limits _0^{2 \pi} cos^2 \theta ~ d \theta = \pi\)


\(\therefore I = \int\limits_{z = 0}^4 \int\limits_{r = 0}^{\sqrt z} \pi r^3 + 2 \pi r z^2 ~ ~dr ~dz \)

\(= \int\limits_{z = 0}^4 \left[ \frac{1}{4}\pi r^4 + \pi r^2 z^2 \right]_{r = 0}^{\sqrt z} ~dz \)

\(= \int\limits_{z = 0}^4 \frac{1}{4}\pi z^2 + \pi z^3 ~dz \)

\(= \left[ \frac{1}{12}\pi z^3 + \frac{1}{4}\pi z^4 \right]_{z = 0}^4 \)

\( = \frac{16}{3} \pi + 64 \pi = \frac{208}{3} \pi \)

Other ways to do it but NOT leaving the dz integration to end unless you switch limits, eg:

\(\int\limits_{z = 0}^{r^2} \int\limits_{r = 0}^{2} \int\limits_{\theta = 0}^{2 \pi} \dots \)

Answer 4

Thank you both c: