# Evaluate Double integral

1. Apr 21, 2008

### tony873004

1. The problem statement, all variables and given/known data
Evaluate the integral $$\int {\int\limits_R {\left( {x + y} \right)\,dA} }$$ where R is the region that lies to the left of the y-axis between the circles $$x^2 + y^2 = 1$$ and $$x^2 + y^2 = 4$$ by changing to polar coordinates.

2. Relevant equations
x=r cos theta
y=r sin theta

3. The attempt at a solution
my effort:

$$\begin{array}{l} r_{inner} = \sqrt 1 = 1,\,\,r_{outer} = \sqrt 4 = 2 \\ R = \left\{ {\left( {r,\theta |1 \le r \le 4,\,\frac{{3\pi }}{2}\,\theta \le \pi } \right)} \right\} \\ x = r\cos \left( \theta \right),\,\,\,y = r\sin \left( \theta \right) \\ \int\limits_1^2 {\int\limits_{\pi /2}^{3\pi /2} {\left( {r\cos \left( \theta \right) + r\sin \left( \theta \right)} \right)\,d\theta \,dr} } \\ \end{array}$$

The solution shows an extra instance of r in the integral. If the original question is for (x+y) then why isn't it simply r cos + r sin, rather than r(r cos + r sin)?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 21, 2008

### rootX

Assuming everything else is correct:

there is an additional r because of Jacobian
you will learn co-od transformations soon
when you transform, area changes, and that
r takes into account of that

3. Apr 21, 2008

### EngageEngage

the area element in polar coordinates is :

$$dA = rdrd\theta$$
that is where the extra r comes from.

4. Apr 21, 2008

### tony873004

Thanks!

I was about to say never mind because I just found the formula in the book. It gives the extra r.

But I'm glad I asked, because now I understand where it comes from.

5. Apr 21, 2008

### EngageEngage

indeed, you can see this also if you draw out the picture of the infinitesimal pieces: rdtheta will be a small slice of length. multiplying by dr will give you approximately a small square. so dxdy =~ rdrdtheta