# Polar Integration Question

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• April 27th 2010, 02:45 AM
Peleus
Polar Integration Question
Hi all,

The question -

Calculate $V = \int \int_R \sin{\sqrt{x^2 + y^2}} dA$

R is the region inside the circle $x^2 + y^2 = 4\pi^2$

My attempt (converting to polar) - $V = \int \int_R z dA$

$z = \sin{\sqrt{x^2 + y^2}}$

Assuming we can find a quarter of the circle, than multiply by 4 for the total area...

$V = 4 \int^{\frac{\pi}{2}}_0 \int^{2\pi}_0 \sin{r} r dr d\theta$

$= 4 \int^{\frac{\pi}{2}}_0 [\sin{r} - r\cos{r}]^{2\pi}_{0} d\theta$

$= 4 \int^{\frac{\pi}{2}}_0 [\sin{2\pi} - 2\pi\cos{2\pi}] d\theta$

Here I'm stuck and quite frankly it's looking a lot more complicated than I imagine it should, can someone please point out the error in my working?

On the off chance I am correct so far, can I treat those $2\pi$ sections as radians? I don't believe I can as that wasn't the context they were used originally, so what would I do?

Thanks for your time folks.
• April 27th 2010, 03:06 AM
Opalg
Quote:

Originally Posted by Peleus
Calculate $V = \int \int_R \sin{\sqrt{x^2 + y^2}} dA$

R is the region inside the circle $x^2 + y^2 = 4\pi^2$

My attempt (converting to polar) - $V = \int \int_R z dA$

$z = \sin{\sqrt{x^2 + y^2}}$

Assuming we can find a quarter of the circle, than multiply by 4 for the total area...

$V = 4 \int^{\frac{\pi}{2}}_0 \int^{2\pi}_0 \sin{r} r dr d\theta$

$= 4 \int^{\frac{\pi}{2}}_0 [\sin{r} - r\cos{r}]^{2\pi}_{0} d\theta$

$= 4 \int^{\frac{\pi}{2}}_0 [\sin{2\pi} - 2\pi\cos{2\pi}] d\theta$

Here I'm stuck and quite frankly it's looking a lot more complicated than I imagine it should, can someone please point out the error in my working?

On the off chance I am correct so far, can I treat those $2\pi$ sections as radians? I don't believe I can as that wasn't the context they were used originally, so what would I do?

Yes, you are correct so far, and $2\pi$ means $2\pi$ whatever context it appears in. So the next step is to say that $\sin2\pi=0$ and $\cos2\pi=1$, and you are just integrating the constant $-2\pi$ with respect to $\theta$. (If you're worried that the answer comes out negative, think what the graph of the function looks like, and you'll see that the answer should indeed be negative.)
• April 27th 2010, 03:13 AM
Peleus
Quote:

Originally Posted by Opalg
Yes, you are correct so far, and $2\pi$ means $2\pi$ whatever context it appears in. So the next step is to say that $\sin2\pi=0$ and $\cos2\pi=1$, and you are just integrating the constant $-2\pi$ with respect to $\theta$. (If you're worried that the answer comes out negative, think what the graph of the function looks like, and you'll see that the answer should indeed be negative.)

Firstly, thank you for your fast reply.

Obviously it's a big confidence boost to know I'm getting something right, it's been a hell of a night (Headbang)(Crying)(Headbang)(Crying)(Headbang)

In regards to the $2\pi$ however, isn't it used in the context of a distance (the radius) rather than radian's for the sin and cos?

So wouldn't $\sin2\pi=0$ and $\cos2\pi=1$ only be correct if it was in fact radians, which from my understanding it is not? For some reason I'm thinking it should be something like $\sin{6.28...} = 0.1094.....$

I'm sure I'm wrong, but it's always important to find out why your wrong, rather than simply getting the right answer (Happy)