Thread: Another Double Integral with Polar Coordinates

1. Another Double Integral with Polar Coordinates

Problem:

Evaluate the iterated integral by converting to polar coordinates.

$\int_{-3}^{3}\int_{0}^{\sqrt{9-x^{2}}}sin(x^{2}+y^{2}) \: dy \:dx$

My attempt:

2. Originally Posted by tangibleLime
Problem:

Evaluate the iterated integral by converting to polar coordinates.

$\int_{-3}^{3}\int_{0}^{\sqrt{9-x^{2}}}sin(x^{2}+y^{2}) \: dy \:dx$

My attempt:
First off, your conversion to polar coordinates is flawless. By the way, there is this nice property that allows us to break apart the integral because the integrand terms are independent of each other. So you can first rewrite the polar integral as: $\left(\int_0^{\pi}\,d\theta\right)\left(\int_0^{3} \sin(r^2)r\,dr\right)$

However, I'm not sure how you did that integration with respect to r, since $\int_0^3\sin(r^2)r\,dr$ requires a substitution..

3. Thanks for the response!

I know about that property, it has definitely saved me some headaches in the past!

I had originally integrated everything by hand, but I arrived at the incorrect end result several times (including this one). I decided to try to isolate my problem by using Wolfram Alpha to do my integration in an attempt to make sure at least the integration portion of the problem was correct.

The exact input string I used was,
int_0^3 (sin(r^2)*r) dr

Which returns $sin^{2}(\frac{9}{2})$.

4. Originally Posted by tangibleLime
Thanks for the response!

I know about that property, it has definitely saved me some headaches in the past!

I had originally integrated everything by hand, but I arrived at the incorrect end result several times (including this one). I decided to try to isolate my problem by using Wolfram Alpha to do my integration in an attempt to make sure at least the integration portion of the problem was correct.

The exact input string I used was,
int_0^3 (sin(r^2)*r) dr

Which returns $sin^{2}(\frac{9}{2})$.
Ohhh...I see now. Because if you made the substitution $u=r^2$, then

\begin{aligned}\int_0^3\sin(r^2)r\,dr \xrightarrow{u=r^2}{}&\tfrac{1}{2}\int_0^{9}\sin(u )\,du\\ =&\tfrac{1}{2}\left.\left[-\cos(u)\right]\right|_0^{9}\\=&\tfrac{1}{2}(1-\cos(9))\end{aligned}

And then they (WolframAlpha) applied the identity $\sin^2x=\tfrac{1}{2}(1-\cos 2x)$ to rewrite $\tfrac{1}{2}(1-\cos(9))$ as $\sin^2(\tfrac{9}{2})$.

So it turns out your answer was right, but now we know why its right.

I hope this makes sense!

5. Ooh, okay, thanks! I had a hunch it had something to do with that identity, but I must have made a mistake when trying to convert it myself.

Thanks again!