# Polar Integration Question

• Apr 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?

• Apr 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.)
• Apr 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.)

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 $\sin{6.28...} = 0.1094.....$