If you guys like to try out a problem. Evaluate triple integral by changing to spherical coordinates \[\large \int\limits_0^a \int\limits_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\left(yx^2+y^3+yz^2\right)dzdxdy\]
yes sure give me easyyyyy one
I am not good at math, but lets see, I might now...
yes
Writing this integral is a little messy, one min.
it is probably going to come out from some derivative.
nope its gonna be chocolates that's why i'm here
you can draw it, if you want.
Nnesha, I will have to disappoint you. There will be no chocolate.
in the question box, refresh.
Wow that's complicated.
No dude, I am out. I have a life
Hahaha
<---------------sorry he/she was right there is no chocolates :((((
I am he.
@Miracrown @ganeshie8 @satellite73 @Concentrationalizing
@Mimi_x3 this si like 1+1 for you :)
@Jhannybean hmmmm :P Im a dwarf
Ok I'll give you guys all a hint
jhanny i have pre calculus next semester so holdon i will help you then okay just wait
Factor out a y from the integrand \[y(\color{red}{x^2+y^2+z^2})dzdydx\]
BIG HINT!!!
Um, idk pi?
What does \(x^2+y^2+z^2\) in spherical coordinates equal? :P
you should make tutorial
Yeah :P More or less.
Nah, I don't have time to make Calc tutorials lmao.
okay you didn't give me chocolates so now i have to eat pizza bye best of luck sorry for spammming
|dw:1418348236393:dw|
Join our real-time social learning platform and learn together with your friends!