# Multiple integrals in polar form

1. May 25, 2012

### robertjford80

1. The problem statement, all variables and given/known data

do you see how the integral of r is .5?

I don't get how that follows?

2. May 25, 2012

### sharks

$$\int^{\theta=\pi}_{\theta=0}\int^{r=1}_{r=0} rdrd\theta$$Then, integrate the inner-most integral and from there, integrate outwards, one integral at a time. So, in this case, integrate w.r.t.r first and then w.r.t.θ.

3. May 25, 2012

### robertjford80

still confused

4. May 25, 2012

### sharks

This is the first step in solving any integral. You should understand which variable relates to the limits, then make the following modification.:
$$\int^{\pi}_{0}\int^{1}_{0} rdrd\theta=\int^{\theta=\pi}_{\theta=0}\int^{r=1}_{r=0} rdrd\theta$$
Next, break the integrals down, starting with the inner-most integral:
$$\int^{r=1}_{r=0} r\,.dr=answer$$
$$\int^{\theta=\pi}_{\theta=0} answer\,.d\theta=final\;answer$$

5. May 25, 2012

### robertjford80

The book says the answer to this

$$\int^{r=1}_{r=0} r\,.dr$$

is .5, I don't get that.

6. May 25, 2012

### robertjford80

Ok, I got it.

the integral of r is r2/2, hence 1/2

7. May 25, 2012

### HallsofIvy

Staff Emeritus
The way I would do this is to note that the integral gives the area between the x-axis and the curve $y= \sqrt{1- x^2}$ from x= -1, to 1. $y= \sqrt{1- x^2} is the part of [itex]y^2= 1- x^2$ or $x^2+ y^2= 1$. That is, it is the area of a semi-circle of radius 1 and so is $\pi/2$.

Of course, $\int_0^\pi d\theta= \pi$ and $\int_0^1 r dr= 1/2$, as you say, so the double integral is $\pi/2$.