2

I'm trying to evaluate this integral

$\displaystyle\int_{-\pi/2}^{\pi/2} d\theta \displaystyle\int_0^{ 2a\cos \theta} r^2 \cos \theta dr $

by reversing the order of integration. Integrating as it stands I can easily get the correct answer $\pi a^3$

However when I try to reverse the order I can't get the same answer

$ \displaystyle \int_{0}^{2a} dr \displaystyle \int_{-\cos ^{-1}(r/(2a))}^{ \cos ^{-1}(r/(2a))} r^2 \cos \theta d \theta = \displaystyle \int_{0}^{2a} r^2 (\frac {2a}{r})(2) dr = 8a^3$ ???

Really banging my head on the wall...anyone care to tell me where I went wrong?

user141554
  • 51

1 Answers1

0

The inner integral in your changed integral isn't just $r^2*(2a/r)*2.$ Integrating $r^2 \cos \theta$ gives $r^2 \sin \theta$, and the limits are inverse cosines. If you then draw the right triangle with hypotenuse $2a$ and $x$ side $r$ you get that $\sin \theta=\sqrt{4a^2-r^2}/(2a).$ So integration of the inner integral gives $1/a*r^2*\sqrt{4a^2-r^2}.$ This when integrated for $r$ from $0$ to $2a$ gives $\pi a^3.$

coffeemath
  • 29,884
  • 2
  • 31
  • 52