Multiple integrals in polar form

[robertjford80]

Homework Statement

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

I don't get how that follows?

 

[DryRun]

First, label your limits:
\int^{\theta=\pi}_{\theta=0}\int^{r=1}_{r=0} rdrd\thetaThen, 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.

 

[robertjford80]

still confused

 

[DryRun]

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

 

[robertjford80]

The book says the answer to this

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

is .5, I don't get that.

 

[robertjford80]

Ok, I got it.

the integral of r is r²/2, hence 1/2

 

[HallsofIvy]

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 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. <br /> <br /> 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.