Register to reply 
Another Polar Double Integral 
Share this thread: 
#1
Nov2406, 07:03 PM

HW Helper
P: 2,685

Hey everyone, My task this time is to derive the volume of a sphere using a polar double integral.
The sphere has radius [tex] a[/tex] we know that r goes from 0 to a in this integral. The equation for a sphere is: [tex] x^2 + y^2 +z^2 = r^2[/tex] or [tex] f(x,y) = \sqrt{r^2 x^2 y^2}[/tex] and it intersects the xy plane in a circle: [tex] x^2 + y^2 = r^2 [/tex] So if we find the volume over this circle and under the positive half of the sphere and double it we should get the volume we want. r goes from 0 to a, and [tex]\theta[/tex] goes from 0 to[tex]2\pi[/tex]. So we get: [tex]2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta [/tex] My problem is that whenever I fill in the polar conversions for x and y my sphere equation becomes z=0. I know I'm missing something simple but I can't figure it out. Any help would be appreciated. Thank you. 


#2
Nov2406, 07:18 PM

Emeritus
Sci Advisor
PF Gold
P: 4,500

Because you're confusing your r's. One of them is a variable, the other is the radius of the sphere



Register to reply 
Related Discussions  
Double integral with polar  Calculus & Beyond Homework  1  
Double integral into the polar form...  Calculus & Beyond Homework  4  
Polar double integral  Calculus & Beyond Homework  7  
Double integral in polar form  Calculus & Beyond Homework  5  
Double integral and polar coord  Calculus  3 