Double integral of polar coordinates

May 2010
74
1
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...

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?
 

Jester

MHF Helper
Dec 2008
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1,255
Conway AR
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...

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?
First, the integral would make more sense if it was written as

\(\displaystyle
\int_a^1 \int_0^{\sqrt{1-x^2}} F(x,y)\,dy\,dx
\)

To switch this to polar

\(\displaystyle
\int_0^{\cos^{-1}a}\int_{a \sec \theta}^1 F(r \cos \theta, r \sin \theta) r dr d \theta
\)
 
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HallsofIvy

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Apr 2005
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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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