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triple integral spherical coordinate

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0

$begingroup$

I have a problem converting this question into a spherical form.
$∫∫∫ z/√(x^2+y^2+z^2)dxdydz$ where R is the interior of a sphere $x^2+y^2+z^2 = 2z$

the limits of integration I found are:
0≤r≤2cosθ
0≤θ≤ π
0≤φ≤2 π

After converting this is my integrand
$∫∫∫ rcosθr^2sinθ/√2rcosθ drdθdφ$ with limit given above.
But this doesn't give me the right answer. Any help will be appreciated. Thanks in advance.

integration multivariable-calculus spheres

share|cite|improve this question

edited Mar 16 at 22:28

josh

asked Mar 16 at 21:50

joshjosh

177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure in the $x^2+y^2+y^2=2z$?
  $endgroup$
  – Botond
  Mar 16 at 22:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, it should be z^2.
  $endgroup$
  – josh
  Mar 16 at 22:28
  

  
  
  
  

add a comment | 

0

$begingroup$

I have a problem converting this question into a spherical form.
$∫∫∫ z/√(x^2+y^2+z^2)dxdydz$ where R is the interior of a sphere $x^2+y^2+z^2 = 2z$

the limits of integration I found are:
0≤r≤2cosθ
0≤θ≤ π
0≤φ≤2 π

After converting this is my integrand
$∫∫∫ rcosθr^2sinθ/√2rcosθ drdθdφ$ with limit given above.
But this doesn't give me the right answer. Any help will be appreciated. Thanks in advance.

integration multivariable-calculus spheres

share|cite|improve this question

edited Mar 16 at 22:28

josh

asked Mar 16 at 21:50

joshjosh

177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure in the $x^2+y^2+y^2=2z$?
  $endgroup$
  – Botond
  Mar 16 at 22:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, it should be z^2.
  $endgroup$
  – josh
  Mar 16 at 22:28
  

  
  
  
  

add a comment | 

$begingroup$

I have a problem converting this question into a spherical form.
$∫∫∫ z/√(x^2+y^2+z^2)dxdydz$ where R is the interior of a sphere $x^2+y^2+z^2 = 2z$

the limits of integration I found are:
0≤r≤2cosθ
0≤θ≤ π
0≤φ≤2 π

After converting this is my integrand
$∫∫∫ rcosθr^2sinθ/√2rcosθ drdθdφ$ with limit given above.
But this doesn't give me the right answer. Any help will be appreciated. Thanks in advance.

integration multivariable-calculus spheres

share|cite|improve this question

edited Mar 16 at 22:28

josh

asked Mar 16 at 21:50

joshjosh

177

$endgroup$

share|cite|improve this question

edited Mar 16 at 22:28

josh

asked Mar 16 at 21:50

joshjosh

177

share|cite|improve this question

edited Mar 16 at 22:28

josh

asked Mar 16 at 21:50

joshjosh

177

asked Mar 16 at 21:50

joshjosh

177

asked Mar 16 at 21:50

joshjosh

177

2 Answers
2

active

oldest

votes

0

$begingroup$

The equation of the sphere can be rewritten as $x^2+y^2+(z-1)^2=1$. Thus the sphere has center $(0,0,1)$ and radius $1$. In particular, it is not the sphere in the picture. Since the sphere lies above the $xy$-plane, you should have
$$0leq theta leq fracpi2.$$
Also the denominator $sqrt2 r cos theta$ in the integrand looks wrong. The denominator is just $r$ since $r = sqrtx^2+y^2+z^2$.

share|cite|improve this answer

edited Mar 16 at 22:59

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

$endgroup$

add a comment | 

0

$begingroup$

Maybe , instead of $$ z=r cos φ $$
$$ z=1+r cos φ $$

Then the boundaries for r would be from 0 to 1

share|cite|improve this answer

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513

$endgroup$

add a comment | 

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0

$begingroup$

The equation of the sphere can be rewritten as $x^2+y^2+(z-1)^2=1$. Thus the sphere has center $(0,0,1)$ and radius $1$. In particular, it is not the sphere in the picture. Since the sphere lies above the $xy$-plane, you should have
$$0leq theta leq fracpi2.$$
Also the denominator $sqrt2 r cos theta$ in the integrand looks wrong. The denominator is just $r$ since $r = sqrtx^2+y^2+z^2$.

share|cite|improve this answer

edited Mar 16 at 22:59

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

$endgroup$

add a comment | 

0

$begingroup$

The equation of the sphere can be rewritten as $x^2+y^2+(z-1)^2=1$. Thus the sphere has center $(0,0,1)$ and radius $1$. In particular, it is not the sphere in the picture. Since the sphere lies above the $xy$-plane, you should have
$$0leq theta leq fracpi2.$$
Also the denominator $sqrt2 r cos theta$ in the integrand looks wrong. The denominator is just $r$ since $r = sqrtx^2+y^2+z^2$.

share|cite|improve this answer

edited Mar 16 at 22:59

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

$endgroup$

add a comment | 

$begingroup$

The equation of the sphere can be rewritten as $x^2+y^2+(z-1)^2=1$. Thus the sphere has center $(0,0,1)$ and radius $1$. In particular, it is not the sphere in the picture. Since the sphere lies above the $xy$-plane, you should have
$$0leq theta leq fracpi2.$$
Also the denominator $sqrt2 r cos theta$ in the integrand looks wrong. The denominator is just $r$ since $r = sqrtx^2+y^2+z^2$.

share|cite|improve this answer

edited Mar 16 at 22:59

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

$endgroup$

share|cite|improve this answer

edited Mar 16 at 22:59

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

answered Mar 16 at 22:50

Ernie060Ernie060

2,795419

0

$begingroup$

Maybe , instead of $$ z=r cos φ $$
$$ z=1+r cos φ $$

Then the boundaries for r would be from 0 to 1

share|cite|improve this answer

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513

$endgroup$

add a comment | 

0

$begingroup$

Maybe , instead of $$ z=r cos φ $$
$$ z=1+r cos φ $$

Then the boundaries for r would be from 0 to 1

share|cite|improve this answer

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513

$endgroup$

add a comment | 

$begingroup$

Maybe , instead of $$ z=r cos φ $$
$$ z=1+r cos φ $$

Then the boundaries for r would be from 0 to 1

share|cite|improve this answer

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513

$endgroup$

share|cite|improve this answer

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513

answered Mar 16 at 23:03

Milan StojanovicMilan Stojanovic

452513