Is that a bit better?Double Integral in Polar Coordinates

[sa1988]

Homework Statement

Homework Equations

The Attempt at a Solution

As with my other recent posts, I just want to check if I'm right or wrong as I don't have an answer scheme to go by.

For this question I simply converted to polar to get:

∫∫(a+a)r drdθ

for 0<r<a, 0<θ<2π ,

which solved to give

2πa³ .

Was that correct? I'm convinced I did something wrong because it seems a very easy 7 marks.

Thanks!

 

[xiavatar]

You almost got it right. You made a silly arithmetical error. The integrand should be (a+r)r not (a+a)r.

$\int_0^a\int_0^{2\pi}(a+r)r d\theta dr$

 

[sa1988]

Aaah right ok. Hmmm maybe a silly arithmetic error, but I can't see why it should be (a+ar)?

For converting to polar I did, dxdy = r drdθ,

which I presumed would then give the expression I used.

I can't see where I went wrong?

 

[xiavatar]

Sorry it should be this $\int_0^a\int_0^{2\pi}(a+r)r d\theta dr$. What is $\sqrt{x^2+y^2}$ after you transform it into polar coordinates? What is $x$ and $y$ in terms of $r$ and $\theta$?

 

[sa1988]

Ah yeah I think I follow now. I think the x² + y² = a² thing threw me a bit, making me think I should just swap it for an 'a' when really I can't do that at all.

It should be

f(x,y)dxdy → f(rcosθ,rsinθ)r drdθ

So in the case of the question

[a + √(x² + y²)] dxdy → [(a + r)r] drdθ