Question

Need help with solving a triple integral.

Answer 1

\[\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\cos(x+y+z)dxdydz\]

Answer 2

If \(u=x+y+z\), then \(\mathrm{d}u=\mathrm{d}x\) and \(0\le x\le \pi\) becomes \(y+z\le u\le\pi+y+z\):
\[\iiint \cos(x+y+z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\int_0^\pi\int_0^\pi\int_{y+z}^{\pi+y+z}\cos u\,\mathrm{d}u\,\mathrm{d}y\,\mathrm{d}z\]The first integral is simple enough:
\[\int_{y+z}^{\pi+y+z}\cos u\,\mathrm{d}u=\sin(\pi+y+z)-\sin(y+z)\]You can use similar substitutions to tackle the rest:
\[\int_0^\pi\int_0^\pi\bigg(\sin(\pi+y+z)-\sin(y+z)\bigg)\,\mathrm{d}y\,\mathrm{d}z\]

Answer 3

this is where I got to.

Answer 4

My Ti-89 says for the first one it's -2sin(y+z)

Answer 5

I think you're going to be disappointed when you're all done :-)

Answer 6

You can actually simplify the last integrand a bit via an identity:
\[\sin(\pi+k)=\sin\pi\cos k+\cos\pi\sin k=-\sin k\]so that the double integral is equivalent to
\[-2\int_0^\pi\int_0^\pi\sin(y+z)\,\mathrm{d}y\,\mathrm{d}z\](I was going to stop there with this post but looks like you beat me to the gist).

Now substitute \(u=y+z\) so that \(\mathrm{d}u=\mathrm{d}y\) and \(0\le y\le\pi\) becomes \(z\le u\le \pi+z\), so the double integral is
\[-2\int_0^\pi\int_z^{\pi+z}\sin u\,\mathrm{d}u\,\mathrm{d}z=-2\int_0^\pi\bigg(\cos z-\cos(\pi+z)\bigg)\,\mathrm{d}z\]

Answer 7

Ohhh so that's how it got that.

Answer 8

That's pretty much all I was having trouble with.

Answer 9

Guess it helps to remember my identities.

Answer 10

Thx.

Answer 11

yw

Answer 12

Answer is 0 btw.

Answer 13

wow all that for a 0 answer. That happened to me in Calculus III
all that computation :/

Answer 14

Lol, happens I guess. You can kind of tell it's going to be 0 though, Because at the end you'll be getting sin(pi) or sin(2pi). Both of which = 0