Answer to Question #142093 in Calculus for Promise Omiponle

Question

Evaluate the iterated integral ∫(0 to π/3)∫(0 to y)∫(0 to x) cos(x+y+z) dzdxdy.

Answer 1

SOLUTION

$\smallint_0^{\frac{\pi}{3}}\int_0^y\int _0^x cos(x+y+z) dzdxdy$

We first consider the inner integral

$=$ $\smallint_0^{\frac{\pi}{3}}\int_0^y [\int _0^x cos(x+y+z) dz]dxdy$

$=\smallint_0^{\frac{\pi}{3}}\int_0^y [ sin(x+y+z)]_0^x dxdy$

$=\smallint_0^{\frac{\pi}{3}}\int_0^y [ sin(x+y+x) - sin(x+y+0)] dxdy$

$=\smallint_0^{\frac{\pi}{3}}\int_0^y [ sin(2x+y) - sin(x+y)] dxdy$

Now we proceed to the middle integral;

$=\smallint_0^{\frac{\pi}{3}}[\int_0^y ( sin(2x+y) - sin(x+y)) dx]dy$

$=\smallint_0^{\frac{\pi}{3}}[ [ cos(x+y) - \frac{cos(2x+y)}{2}]_0^y]dy$

$=\smallint_0^{\frac{\pi}{3}}[ \frac{-cos(3y)-2cos(2y)+cos(y)}{2}]dy$

And now taking the outer integral, we have;

$\smallint_0^{\frac{\pi}{3}}[ \frac{-cos(3y)-2cos(2y)+cos(y)}{2}]dy$

$=[ \frac{- \frac{sin(3y)}{3}+sin(2y)-sin(y) }{2}]_0^{\frac{\pi}{3}}$

$=[ \frac{- \frac{sin(\pi)}{3}+sin(\frac{2}{3}\pi)-sin(\frac{\pi}{3}) }{2} - (\frac{- \frac{sin(0)}{3}+sin(0)-sin(0) }{2})]$

$=[ \frac{- 0+0.866-0.866 }{2} - 0)]$

$=0-0$

$=0$