# Changing order of integration (double) in polar coord

1. Sep 5, 2013

### nothingness00

1. The problem statement, all variables and given/known data

Change the order of the limits of integration of the following double integral and evaluate.

2. Relevant equations

$\int_{0}^\frac{\pi}{2} \int_{0}^{cos(\theta)} cos(\theta)\,dr\,d\theta$

3. The attempt at a solution

Evaluating as it is, I arrive an answer of $\frac{\pi}{4}$.

I know the region to be integrated is the semicircle bounded by the polar axis, with corner points at r = 0, and r = 1, with a height of 1/2. I know that normally, in the cartesian case, to change the order of integration requires the limits to be written from y(x) to x(y), with the x or y intervals adjust accordingly.

Thus, for this problem, the original region is bounded by:

$0<r<cos(\theta)$ and $0<\theta<\frac{\pi}{2}$.

Changing the form, I would write,

$0<\theta<cos^{-1}(r)$ and $0<r<1$

Trying to evaluate in this manner, I end up at

$\int_{0}^1 sin(cos^{-1}(r))\,dr$,

after which I cannot go further. I have very little experience of changing limit orders in the polar case. Any hints would be appreciated!

Last edited: Sep 5, 2013
2. Sep 5, 2013

### CompuChip

Hint: note that sin x = sqrt ( 1 - cos^2 x ).

3. Sep 5, 2013

### nothingness00

Cool, thanks; Using the hint: I get

$\int_{0}^1 \sqrt{1-cos(r)}\,dr$

Multiplying by a factor of sin(r)/sin(r), I get,

$\int_{0}^1 \frac{sin(r)}{\sqrt{1+cos(r)}}\,dr$.

I use a substitution of

$u = 1 + cos(r)$ with
$\,du = -sin(r)\,dr$.

I substitute the limits by using

$u_0 = 1+cos(0) = 2$
$u_1 = 1+cos(1) = ~1.5403...$

Evaluating the new integral, through

$-[2u^{\frac{1}{2}}]$

between u1=1.5403... and u0 = 2, I get an answer of ~0.173144 which is not quite $\frac{\pi}{4}$

Are my limits wrong, or did I blunder somewhere else?

4. Sep 6, 2013

### CompuChip

Unfortunately the first line is still incorrect.

What would x be, looking at your integrand (it's not just r, as you seem to have assumed!).