Volume using double integral and polar coordinates

  • #1
Find the volume under the cone z = sqrt ( x2+y2 ) and on the disk x2+y2 < 4. Use polar coordinates.

Graphing x2+y2 < 4, I get a circle centered at 0,0 with radius of 2
So theta goes from 0 to 2pi

Also, since x2+y2 < 4
This means that r^2 < 4
so -2 < r < 2

[tex]\int_{0}^{2\pi}\int_{-2}^{2}r\sqrt{x^{2}+y^{2}} dr d\theta=\int_{0}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta
= \frac{32\pi}{3}[/tex]
but the answer is 16pi/3

Can anyone help me
 

Answers and Replies

  • #2
radou
Homework Helper
3,115
6
Your integration boundaries for r are not correct.
 
  • #3
cronxeh
Gold Member
961
10
radius can't be negative, its length is still 2
[tex]\int_{0}^{2\pi}\int_{0}^{2}r^{2} dr d\theta [/tex]

(although if you negate the radius, you increase theta by pi):
[tex]\int_{pi}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta [/tex]

(probably just easier to think of radius as a non-negative variable)
 
Last edited:

Related Threads on Volume using double integral and polar coordinates

Replies
3
Views
3K
Replies
2
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
7K
Replies
18
Views
4K
Replies
0
Views
1K
Top