MHB 15.1.34 Evaluate triple integral

[karush]

15.1.34 Evaluate
$\displaystyle I=\int_{0}^{3\pi/2}\int_{0}^{\pi}\int_{0}^{\sin{x}}
\sin{y} \, dz \, dx \, dy$
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

 

[tkhunny]

Show the detail on that dx piece.

 

[HallsofIvy]

15.1.34 Evaluate
$\displaystyle I=\int_{0}^{3\pi/2}\int_{0}^{\pi}\int_{0}^{\sin{x}}
\sin{y} \, dz \, dx \, dy$
integrat dz
$\displaystyle I=\int _0^{3\pi/2}\int _0^{\pi }\sin(y)\sin (x)\, dxdy $
integrat dx$

This is correct and it can be written
$I=\left(\int_0^{3\pi/2} sin(y)dy\right)\left(\int_0^\pi sin(x)dx\right)$
$= \left[-cos(y)\right]_0^{3\pi/2}\left[-cos(x)\right]_0^\pi= (1)(2)= 2$

By the way, in English, "integrate" has an "e" on the end.

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

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