Question

HELP!!! Evaluate this triple integral.....

Answer 1

What is it?

Answer 2

\[\int\limits_{0}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\pi}\int\limits_{0}^{2}e ^{-p ^{3}}p ^{2}dpd \theta d \phi \]

Answer 3

Well, the integrals over theta and phi are easy. You don't have any dependence on them at all.

For the integral over p, choose u=p^3 and see what you come up with ;)

Answer 4

why not u=-p^3?

Answer 5

That will also work

Answer 6

so do I take a -3 out of the integral?

Answer 7

Not quite.

\[u = -p^3\]
\[du = -3p^2dp\]

So:
\[p^2dp = \frac{-1}{3}du\]

Answer 8

so I would have this for p?\[-\frac{ 1 }{ 3 }\int\limits_{0}^{2}e ^{u}du\]

Answer 9

Yes, but your bounds aren't 0 to 2 anymore (unless you substitute back in for u after doing the integral)

Answer 10

How do I find the new bounds?

Answer 11

u = -p^3

Your bounds are values for p.
Plug them in and see what your new bounds for u are!

Answer 12

-8 and 0?

Answer 13

Yep :)

Answer 14

So would the bounds be this?\[\int\limits_{0}^{-8}\]

Answer 15

yeah

Answer 16

ok so after the integral is it (e^-8)-1?

Answer 17

Yep.

\[\int\limits^{-8}_0e^udu = e^{-8} - 1\]

You had a -1/3 somewhere, as well, don't forget. And you need to do the other two integrals (very easy)

Answer 18

Should I multiply the -1/3 now or after the two integrals?

Answer 19

Doesn't matter. The result from the other two integrals is multiplied to this, as well, as this doesn't have a theta or phi in it.

Answer 20

So after the second integral would it be\[e ^{-8} \theta - \theta\] bounds 0 to pi?

Answer 21

yeah. You may want to just do it as:
\[(e^{-8}-1)\theta\], just to save some time simplifying later.

Answer 22

Ok so the final integral setup would be?\[-\frac{ 1 }{ 3 }\int\limits_{0}^{\frac{ \pi }{ 2 }}(e ^{-8}-1)\pi d phi \]

Answer 23

Yep

Answer 24

So would it be?\[-\frac{ 1 }{ 3 }(e ^{-8}-1)\pi \phi \]from 0 to pi/2?

Answer 25

yep

Answer 26

So is the final answer this?\[-\frac{ \pi ^{2} }{ 6 }(e ^{-8}-1)\]

Answer 27

Yep :)

You could absorb the negative into the (e-1) term by switching the order, if you wanted, but what you have is correct.

Answer 28

Good work :)

Answer 29

So when you mean by absorbing the negative, they both cancel?

Answer 30

\[\frac{-\pi^2}{6}(e^{-8}-1) = \frac{\pi^2}{6}(1-e^{-8})\]

Answer 31

Ah ok, thanks for your help, I really appreciate it!! :)

Answer 32

My pleasure :)