a. ∫∫d (x+y)dA wehre D is bound by the X axis, y- axis, and circles x^{2}+y^{2}=4 and x^{2}+y^{2}=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

Printable View

- Apr 8th 2012, 06:11 PMmridgwaSolve double integral in Polar or Cylindrical Coordinates
a. ∫∫d (x+y)dA wehre D is bound by the X axis, y- axis, and circles x

^{2}+y^{2}=4 and x^{2}+y^{2}=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 PMProve ItRe: Solve double integral in Polar or Cylindrical Coordinates
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

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