Question

iteriated integral

Answer 1

where

Answer 2

??

Answer 3

\[\int\limits_{0}^{pi/2}\int\limits_{0}^{\cos \theta}r^2\sin^2\theta drd \theta\]

Answer 4

I think the first one is straightforward. For the second one, how about writing the cos^2 as (1-sin^2)? Then u = sinx, du = cosxdx

Answer 5

alright so lets do the inside first with respect to r and that is (r^3/3)sintheta^2 between o and cos theta lets use fundamental theorem of calculus to get 1/3(cosx)^3(sinx)^2 and now to to integrate this in respect to x

Answer 6

yeah @bmp is right follow that substitution you willl end up with your answer

Answer 7

i was having a serious brain freeze, thanks for the the ideas

Answer 8

No problem :-)

Answer 9

okay i got \[\int\limits_{0}^{pi/2}\cos^3\theta/3 (\sin^2\theta)\]

Answer 10

then i take out the 1/3and i get:\[1/3\int\limits_{0}^{pi/2}(1-\sin^2\theta)\sin^2\theta(\cos \theta)\]

Answer 11

then letting u=sintheta i get:

\[1/3\int\limits_{0}^{pi/2}\sin^2\theta-\sin^4\theta\]

Answer 12

is this correct?

Answer 13

make that sin to u that was the point of substitution right

Answer 14

so you have u^2 -u^4 then integrate that easily with respect to u THEN sub your sin theta back in for u

Answer 15

yeah, i just caught that:\[1/3\int\limits_{0}^{pi/2}u^2-u^4\]

Answer 16

thanks