- #1

karush

Gold Member

MHB

- 3,269

- 5

$\displaystyle I=\int_{0}^{3\pi/2}\int_{0}^{\pi}\int_{0}^{\sin{x}}

\sin{y} \, dz \, dx \, dy$

integrat dz

$\displaystyle I=\int _0^{3\pi/2}\int _0^{\pi }\sin(y)\sin (x)\, dxdy $

integrat dx

$\displaystyle I=\int _0^{3\pi/2}\sin \left(y\right)\cdot \,2dy$

integrat dy

$I=2$

ok I think its correct, took me an hour to do... trig was confusing

typos mabybe