# Math Help - multiple integral

1. ## multiple integral

I have a question about evaluating the iterated integral

its a homework problem i was doing ok with the double integrals but the triple confuse me

int(int(int(x*cosy,x,y,z,0,4,0,pi/2,0,1-x)))

I think that is how you right it well i know it the problem but i was trying to get the signs to actually come up

2. It should look like this:

$\int_0^4\int_0^{\frac{\pi}{2}}\int_0^{1-x}xcosy~dx~dy~dz$

Verify that this is the right integral and I'll start working on it.

3. Thank you !!!!

4. yes this is correct!!!

5. Well, here's my work so far:

I have:

$\int_0^{1-x} xcosy~dx = \frac{1}{2}x^2cos(y)]_0^{1-x}$

$=\frac{1}{2}(1-x)^2cos(y)$

And then:

$\int_0^{\frac{\pi}{2}} \frac{1}{2}(1-x)^2cos(y)~dy$

$=\frac{1}{2}(x-1)^2sin(y)]_0^{\frac{\pi}{2}}$

$= \frac{1}{2}(x-1)^2$

And then:

$\int_0^4 \frac{1}{2}(x-1)^2~dz$

$=\frac{z}{2}(x-1)^2]_0^4$

$=\frac{4}{2}(x-1)^2 = 2(x-1)^2$

6. ## Dont understand

But i dont get it because some how the answer is supposed to be -40/3
and if you have already evaluated it from o to 4 how is that possible

7. $\int_0^4\int_0^{\frac{\pi}{2}}\int_0^{1-x}xcosy~dz~dy~dx$

I think the order of integration got messed up
we need to do z first

$\int_0^4\int_0^{\frac{\pi}{2}}zxcos|_{0}^{1-x}~dy~dx$

$\int_0^4\int_0^{\frac{\pi}{2}}(x-x^2)cos(y)~dy~dx$

$\int_0^4(x-x^2)sin(y)|_{0}^{\pi/2}~dx$

$\int_{0}^{4}x-x^2~dx=\frac{1}{2}x^2-\frac{1}{3}x^3|_{0}^{4}=8-\frac{64}{3}=-\frac{40}{3}$