# Solve double integral in Polar or Cylindrical Coordinates

• Apr 8th 2012, 06:11 PM
mridgwa
Solve double integral in Polar or Cylindrical Coordinates
a. ∫∫d (x+y)dA wehre D is bound by the X axis, y- axis, and circles x2+y2=4 and x2+y2=16 and is in the first quadrant

b. ∫∫∫E Z*sqrt(x^2+y^2) dV over the cylinder x^2+y^2<=4 FOR 1<=z<+5
• Apr 8th 2012, 08:46 PM
Prove It
Re: Solve double integral in Polar or Cylindrical Coordinates
Quote:

Originally Posted by mridgwa
a. ∫∫d (x+y)dA wehre D is bound by the X axis, y- axis, and circles x2+y2=4 and x2+y2=16 and is in the first quadrant

b. ∫∫∫E Z*sqrt(x^2+y^2) dV over the cylinder x^2+y^2<=4 FOR 1<=z<+5

In the first quadrant, the quarter circles have radii 2 and 4 units respectively, and are both swept out between the angles \displaystyle \displaystyle \begin{align*} 0 \leq \theta \leq \frac{\pi}{2} \end{align*}. After making the transformation \displaystyle \displaystyle \begin{align*} x = r\cos{\theta} \end{align*} and \displaystyle \displaystyle \begin{align*} y = r\sin{\theta} \end{align*} and noting that \displaystyle \displaystyle \begin{align*} dA = dx\,dy \to r\,dr\,d\theta \end{align*} the double integral becomes

\displaystyle \displaystyle \begin{align*} \int{\int_D{x + y}\,dA} &= \int_0^{\frac{\pi}{2}}{\int_2^4{\left(r\cos{\theta } + r\sin{\theta}\right)r\,dr}\,d\theta} \\ &= \int_0^{\frac{\pi}{2}}{ \int_2^4{ r^2 \left( \cos{\theta} + \sin{\theta} \right ) \,dr } \, d\theta} \end{align*}