Question

Integrate
∫0to2pi∫1/4to1−r3(2+sin2θ)+rdrdθ

Answer 1

\[\int\limits_{0}^{2pi}\int\limits_{(1/4)}^{1}(-r^3(2+\sin(2\theta))+r)rdrd \theta\]

Answer 2

First compute the inner integral, treating theta as a constant.

Answer 3

whooops forget the last r it shld just br drdtheta
i wish i can edit this thing

Answer 4

It surely looks quite ugly. What did you get as answer?

Answer 5

not telling u :P

Answer 6

I got \[- \frac{15}{1024} (17\sin(\theta) + 2)\]Could someone check my answer? So, I guess it's quite ugly indeed.

Answer 7

For the first integral, I mean.

Answer 8

it's just ugly not hard
ok let's see if I get what bmp gets...

Answer 9

And it should be sin(2theta)

Answer 10

so are you asking this question?\[\huge\int\limits\limits\limits_{0}^{2 \pi}\int\limits\limits\limits_{(1/4)}^{1}(-r^3(2+\sin(2\theta))+r)drd \theta\]

Answer 11

yes i am ya i messed up when typing it out

Answer 12

well wolfram got the same thing as u @bmp

Answer 13

for the first integral

Answer 14

so then just put the second one into wolfram, who really wants to evaluate it?

Answer 15

lol i agreeeeeee ya like wolfram just messes up sometimes with these ugly stuff like everytime i put it in i get diff answers so ill just compute the rest

Answer 16

\[- \frac{15}{1024} \int_{0}^{2\pi}(17\sin(\theta) + 2)d\theta=- \frac{15}{1024} [-17\cos\theta+2\theta]_{0}^{2\pi}=-\frac{60\pi}{1024}\]did I mess up?

Answer 17

NO that is correct yay

Answer 18

-0.6978... I found something like that..

Answer 19

like i got the answer in teh first place but liek I always doubt myself idk

Answer 20

fair enough, I can always use the practice :)

Answer 21

haha thanks guys for ur help