1. Apr 26, 2010 #1

   fishingspree2

   

   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

    

   fishingspree2, Apr 26, 2010
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3. Apr 26, 2010 #2

   radou

   

   Homework Helper

   Your integration boundaries for r are not correct.

    

   radou, Apr 26, 2010
4. Apr 26, 2010 #3

   cronxeh

   

   Gold Member

   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: Apr 26, 2010

   cronxeh, Apr 26, 2010

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