Finding limits of integral in spherical coordinates

uzman1243
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Homework Statement


The question asks me to convert the following integral to spherical coordinates and to solve it
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Homework Equations

The Attempt at a Solution


just the notations θ = theta and ∅= phi

dx dy dz = r2 sinθ dr dθ d∅
r2 sinθ being the jacobian

and eventually solving gets me
∫ ∫ ∫ r4 *sin2θ * sin∅ dr dθ d∅

How do I find the limits now?
 

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Use the limits in the Cartesian system to figure out the enclosed shape. What is the minimum and maximum value of z? Those of y and x?
 
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ehild said:
Use the limits in the Cartesian system to figure out the enclosed shape. What is the minimum and maximum value of z? Those of y and x?

x goes from -2 to 2
y goes from 0 to √4-x2 circle with radius 2
z from 0 to √4-x2-y2 sphere with with radius 2

so I am guessing
r goes from 0 to 2
∅ and θ from 0 to 2π
 
Are you sure in 2pi?
 
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ehild said:
Are you sure in 2pi?
ok so ∅ goes from 0 to 2pi as it is some sort of sphere/ elipse. correct?
 
See picture. Yes, the shape is spherical, but you have to integrate with respect to y from zero to some positive value, goes it round a whole circle?

intshape.JPG
 
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ehild said:
See picture. Yes, the shape is spherical, but you have to integrate with respect to y from zero to some positive value, goes it round a whole circle?

View attachment 82540
ahhh so θ goes from 0 to pi
∅ goes from 0 to 2pi
and r goes from 0 to 2
correct?
 
Those were the limit for the whole sphere. But the integration does not go for negative z values, neither for negative y values.
 
ehild said:
Those were the limit for the whole sphere. But the integration does not go for negative z values, neither for negative y values.
r goes from 0 to 2
theta goes from 0 to pi/2
phi goes from 0 to pi
correct?
 
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Looks good.
 
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uzman1243 said:
r goes from 0 to 2
theta goes from 0 to pi/2
phi goes from 0 to pi
correct?
Yes. :smile:
 
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