Cylindrial Coordinates to evaluate the integral

nysnacc
Messages
184
Reaction score
3

Homework Statement


upload_2016-9-27_22-24-29.png


Homework Equations


x^2+y^2 = r^2

The Attempt at a Solution


I think it is asking me to find the volume of the sphere, which is in the first octant (1/8 of the sphere)

So I set
0<= z<= √1-r2
0 <= r <= 1
0<=θ<=π/2
 
Physics news on Phys.org
Hi, remember that if you want the volume of ##D## you must calculate ##\int\int\int_{D}1dxdydz## where ##f(x,y,z)=1## ...
 
I think the question wants me to use the du dv
 
nysnacc said:
I think it is asking me to find the volume of the sphere

no because ##f(x,y,z)\not=1## ...
 
?
 
What I want to say is that your coordinates are ok ## x=r\cos{\theta}, y=r\sin{\theta}, z=z ## but evaluating the integral you don't have the volume of the sphere (as you said ...) but simply a result that is the evaluation of the integral ...
 
So simply put, do I just plug in the limits:
0<= z<= √1-r2
0 <= r <= 1
0<=θ<=π/2

and into the zex2+y2+z2

then the integral will be something like ∫∫∫∫ z dz r dr dθ
 
nysnacc said:
So simply put, do I just plug in the limits:
0<= z<= √1-r2
0 <= r <= 1
0<=θ<=π/2

and into the zex2+y2+z2

then the integral will be something like ∫∫∫∫ z dz r dr dθ
Your integrand will be ##ze^{x^2+y^2+z^2}## in cylindrical coordinates with ##dV = r~dz~dr~d\theta##.
 
  • Like
Likes Ssnow
LCKurtz said:
Your integrand will be ##ze^{x^2+y^2+z^2}## in cylindrical coordinates with ##dV = r~dz~dr~d\theta##.

Okay but where does z goes (it was before ##e^{x^2+y^2+z^2}##)
 
  • #10
nysnacc said:
Okay but where does z goes (it was before ##e^{x^2+y^2+z^2}##)
?. Right where I have it.
 
  • #11
I rewrite my thought
Originally I have limits:
0<= z<= √1-r2
0 <= r <= 1
0<=θ<=π/2

Then I use them on ∫ ∫ ∫

replace also ex^2+y^2+z^2 with er2+z2

Then I have ∫ ∫ ∫ z er2+z2 r dz dr dθ
 
Back
Top