Ac3
  • Ac3
Need help with solving a triple integral.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Ac3
  • Ac3
\[\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\cos(x+y+z)dxdydz\]
anonymous
  • anonymous
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: \[\int_0^\pi\int_0^\pi\bigg(\sin(\pi+y+z)-\sin(y+z)\bigg)\,\mathrm{d}y\,\mathrm{d}z\]
Ac3
  • Ac3
this is where I got to.

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Ac3
  • Ac3
My Ti-89 says for the first one it's -2sin(y+z)
whpalmer4
  • whpalmer4
I think you're going to be disappointed when you're all done :-)
anonymous
  • anonymous
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 \[-2\int_0^\pi\int_z^{\pi+z}\sin u\,\mathrm{d}u\,\mathrm{d}z=-2\int_0^\pi\bigg(\cos z-\cos(\pi+z)\bigg)\,\mathrm{d}z\]
Ac3
  • Ac3
Ohhh so that's how it got that.
Ac3
  • Ac3
That's pretty much all I was having trouble with.
Ac3
  • Ac3
Guess it helps to remember my identities.
Ac3
  • Ac3
Thx.
anonymous
  • anonymous
yw
Ac3
  • Ac3
Answer is 0 btw.
UsukiDoll
  • UsukiDoll
wow all that for a 0 answer. That happened to me in Calculus III all that computation :/
Ac3
  • Ac3
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

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