I have an intergral and im not sure how to find the limits of integration from cartesian to polar form 0<a<1

Integral from 1 to a ( integral from square root of (1-x^2) to 0) of F(x,y) dydx

Apparently the polar form integrals are suppose to be:

Integral from 1 to a (integral from cos-1(a/r) to 0) and im not sure where that came from...

You find the limits of integration

**geometrically**.

\(\displaystyle \int_{x=1}^a\int_{\sqrt{1-x^2}}^0 F(x,y)dydx\)

means that s ranges from 1 to a and, for each x, y ranges from \(\displaystyle \sqrt{1- x^2} to 0. \(\displaystyle y= \sqrt{1- x^2}\) leads to \(\displaystyle x^2+ y^2= 1\) and, since y is the non-negative root, that is the upper half of the unit circle. Since \(\displaystyle 0\le \sqrt{1- x^2}\) for all x between -1 and 1 (and not defined otherwise), that integral is from the

**upper** curve to the lower. That's why Danny said it would make more sense to reverse the limits of integration on the y integral. Of course, that multiplies the integral by -1: \(\displaystyle -\int_{x= a}^1\int_{y= 0}^{\sqrt{1- x^2}} F(x,y)dydx\).

Now, x= 1 is the right end of that semi-circle while x= a, for a some number between -1 and 1. Points on the unit circle can be written \(\displaystyle (x, y)= (cos(\theta), sin(\theta))\) (because then \(\displaystyle x^2+ y^2= sin^2(\theta)+ cos^2(/theta)= 1\)) so if \(\displaystyle x= cos(\theta)= a\), \(\displaystyle \theta= cos^{-1}(a)\), the arccosine. However, I disagree that the original integral is the same as what you give. If, for example, -1< a< 0, the left side of the region will be vertical line x= a, from y= 0 to \(\displaystyle \sqrt{1- x^2}\). For \(\displaystyle \theta\) from \(\displaystyle cos^{-1}(a)\) to \(\displaystyle \pi\), r will range from 0 to the distance to that vertical line, which is \(\displaystyle \frac{a}{cos(\theta)}\). For \(\displaystyle \theta= 0\) to \(\displaystyle cos^{-1}(a)\) r= 1.

So the integral should be \(\displaystyle \int_{\theta= 0}^{cos^{-1}(a)}\int_{r= 0}^1 F(rcos(\theta),rsin(\theta)) r dr d\theta\)\(\displaystyle + \int_{\theta= cos^{-1}(a)}^\pi \int_{r= 0}^{\frac{a}{cos(\theta)}}F(rcos(\theta), rsin(\theta)) r dr d\theta\).

Also if i have an integral from 1 to a(integral from root(1-x^2) to 0) of (x^2 -y^2)dydx i change it to polar form

integral 1 to a( integral from cos-1(a/r) of r(r^2cos^2theta - r^2sin^2thetha)dtheta dr

I attempted this integral by integrating with respect to r first then in respect to theta but am not sure if its right..any tips?

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