Triple integral using cylindrical coordinates

  • Thread starter jonnyboy
  • Start date
  • #1
18
0

Homework Statement


[tex]\int\int_{Q}\int(x^4+2x^2y^2+y^4)dV[/tex] where Q is the cylindrical solid given by [tex]\{(x,y,x)| x^2+y^2 \leq a^2, 0\leqz\leq\frac{1}{\pi}\}[/tex]


Homework Equations



When I convert to cylindrical I get [tex] f(r,\theta,z) = r^4\cos^2\theta + 2r^4\cos^2\theta\sin^2\theta + r^2\sin^2\theta[/tex], but I just need the bounds for dr, is it? [tex]\int^a_0[/tex]

The Attempt at a Solution


[tex]\int^2\pi_0\int^a_0\int^\frac{1}{\pi}_0 f(r,\theta,z)rdzdrd\theta[/tex]
*the first integral is supposed to be from 0 to 2pi
 
Last edited:

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Your expression for [itex]f(r,\theta,z)[/itex] is incorrect. It might help you to realize that [itex]f(x,y,z)=x^4+2x^2y^2+y^4=(x^2+y^2)^2[/itex].....so [itex]f(r,\theta,z)=?[/itex] :wink:

And yes, [itex]r[/itex] goes from 0 to a.
 
  • #3
18
0
Got it, so [tex]f(r,\theta,z) = r^4[/tex] Much simpler!
 

Related Threads on Triple integral using cylindrical coordinates

Replies
3
Views
2K
Replies
8
Views
2K
Replies
3
Views
3K
Replies
1
Views
1K
Replies
7
Views
7K
Replies
5
Views
6K
Replies
4
Views
2K
Replies
9
Views
4K
Replies
2
Views
729
Replies
1
Views
725
Top