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tdabboud
 3 years ago
iterated integral
tdabboud
 3 years ago
iterated integral

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tdabboud
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/2} \int\limits_{0}^{\cos \theta} r^3 drd \theta\]

polpak
 3 years ago
Best ResponseYou've already chosen the best response.0So integrate r first. What is \[\int_0^{cos \theta} r^3 dr\]

tdabboud
 3 years ago
Best ResponseYou've already chosen the best response.0Thats the thing I am getting stuck right after that so I have this: \[\int\limits_{0}^{\pi/2}[[\cos \theta]^4/4  1/4] d \theta\]

tdabboud
 3 years ago
Best ResponseYou've already chosen the best response.0taking the integral of that before evaluating seems to be some long number however

chaguanas
 3 years ago
Best ResponseYou've already chosen the best response.0I think you need a little more work. cos theta shouldn't be an end point. As part of the set up you should evaluate cos theta and come up with some number for example cos pi/4 is square root 2

polpak
 3 years ago
Best ResponseYou've already chosen the best response.0It should be just \[\int_0^{\pi/2}[\frac{1}{4}cos^4\theta ]d\theta\]

polpak
 3 years ago
Best ResponseYou've already chosen the best response.0Evaluating \[\frac{r^4}{4}\] at r=0 does not give you 1/4

tdabboud
 3 years ago
Best ResponseYou've already chosen the best response.0wow I am dumb, I was evaluating the cos again, okay you were right with the last statement

polpak
 3 years ago
Best ResponseYou've already chosen the best response.0And for integrating cos^4 I think you need to make use of the half angle formula.

tdabboud
 3 years ago
Best ResponseYou've already chosen the best response.0or could you just uses u substitution?

polpak
 3 years ago
Best ResponseYou've already chosen the best response.0You can try, but I don't think it'll work out as nicely.

tdabboud
 3 years ago
Best ResponseYou've already chosen the best response.0okay well thanks for your help atleast I got passed that one error I should be able to get it from here
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