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Calculating a volume and the centroid of a set $S$

Calculating centre of mass of a cylindrical wedgeHaving trouble setting up the limits of integration.Volume of the solid between these two parabaloids - can someone verify my answer?Convert the integral from rectangular to cylindrical coordinates and solveHow to find centroid of this region bounded by surfacesConversion from Cartesian to spherical coordinates, calculation of volume by triple integrationVolume between two surfaces?Help in finding $z$ coordinate of centroid by triple integrationtriple integrals and cylindrical coordinatesCalculate integral where S is the surface of the half ball $x^2 + y^2 + z^2 = 1, space z geq 0,$ and $F = (x + 3y^5)i + (y + 10xz)j + (z - xy)k$

0

$begingroup$

$S$ is the region that is enclosed by the unit sphere $x^2+y^2+z^2=1$ and the cone $z=(x^2+y^2)^1/2$.

So I decided to use cylindrical coordinates first:

Define $G(r,theta,z)=(rcostheta,rsintheta, z)$ and $det DG=r$

This means our first condition becomes $r^2+z^2=1$ and second one becomes $z=r$.

$Volume(S)=intintint_SrdV$. I am not sure how to find the limits of integration.

I know $thetain[0,2pi]$. And we know $z=r implies 2r^2=1implies r= pmfrac1sqrt2 implies -frac1sqrt2leq r leqfrac1sqrt2...?$ And $z=pmsqrt1-r^2?$ I'm not sure if it's supposed to be the other way around for $r$ and $z$ and how are you supposed to know?

The mass is defined to be $intintint_S rho(x)dV$ but I'm not sure what's $rho(x)$... But I know how to calculate the centroid.

calculus multivariable-calculus volume

share|cite|improve this question

asked Mar 15 at 4:43

javacoderjavacoder

848

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You do realize this is just a spherical sector, the volume of which can be calculated by $V_K=fracpi3 a^2 (r-h)$ ?
  $endgroup$
  – Max
  Mar 15 at 9:08
  

  
  
  
  

add a comment | 

0

$begingroup$

$S$ is the region that is enclosed by the unit sphere $x^2+y^2+z^2=1$ and the cone $z=(x^2+y^2)^1/2$.

So I decided to use cylindrical coordinates first:

Define $G(r,theta,z)=(rcostheta,rsintheta, z)$ and $det DG=r$

This means our first condition becomes $r^2+z^2=1$ and second one becomes $z=r$.

$Volume(S)=intintint_SrdV$. I am not sure how to find the limits of integration.

I know $thetain[0,2pi]$. And we know $z=r implies 2r^2=1implies r= pmfrac1sqrt2 implies -frac1sqrt2leq r leqfrac1sqrt2...?$ And $z=pmsqrt1-r^2?$ I'm not sure if it's supposed to be the other way around for $r$ and $z$ and how are you supposed to know?

The mass is defined to be $intintint_S rho(x)dV$ but I'm not sure what's $rho(x)$... But I know how to calculate the centroid.

calculus multivariable-calculus volume

share|cite|improve this question

asked Mar 15 at 4:43

javacoderjavacoder

848

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You do realize this is just a spherical sector, the volume of which can be calculated by $V_K=fracpi3 a^2 (r-h)$ ?
  $endgroup$
  – Max
  Mar 15 at 9:08
  

  
  
  
  

add a comment | 

$begingroup$

$S$ is the region that is enclosed by the unit sphere $x^2+y^2+z^2=1$ and the cone $z=(x^2+y^2)^1/2$.

So I decided to use cylindrical coordinates first:

Define $G(r,theta,z)=(rcostheta,rsintheta, z)$ and $det DG=r$

This means our first condition becomes $r^2+z^2=1$ and second one becomes $z=r$.

$Volume(S)=intintint_SrdV$. I am not sure how to find the limits of integration.

I know $thetain[0,2pi]$. And we know $z=r implies 2r^2=1implies r= pmfrac1sqrt2 implies -frac1sqrt2leq r leqfrac1sqrt2...?$ And $z=pmsqrt1-r^2?$ I'm not sure if it's supposed to be the other way around for $r$ and $z$ and how are you supposed to know?

The mass is defined to be $intintint_S rho(x)dV$ but I'm not sure what's $rho(x)$... But I know how to calculate the centroid.

calculus multivariable-calculus volume

share|cite|improve this question

asked Mar 15 at 4:43

javacoderjavacoder

848

$endgroup$

share|cite|improve this question

asked Mar 15 at 4:43

javacoderjavacoder

848

share|cite|improve this question

asked Mar 15 at 4:43

javacoderjavacoder

848

asked Mar 15 at 4:43

javacoderjavacoder

848

asked Mar 15 at 4:43

javacoderjavacoder

848

1 Answer
1

active

oldest

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0

$begingroup$

The limits for $theta$ are correct. For $r$, as it has to be positive, are $0leq rleqdfrac1sqrt2$. Now, the region of integration is an inverted cone with its vertex at the origin (the equation considers only the positive square root) and the sphere as its "ceiling", so, we have different formulas for the lower limit and for the upper one: $rleq zleq sqrt1-r^2$

$$V(S)=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2rmathbb dz,mathbb dr,mathbb dtheta$$

For the centroid we have to evaluate three integrals, one for each coordinate $C=(C_r,C_theta,C_z)$ but we can take advantage of the region's symmetry of rotation around the $z$ axis. The centroid is in some point of this axis $C=(0,0,C_z)$, with,

$$C_z=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2zrmathbb dz,mathbb dr,mathbb dtheta$$

share|cite|improve this answer

edited Mar 15 at 19:21

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can't understand why the lower limit for $z$ is $r$... How do you know? I understand the lower and upper limit for $r$.
  $endgroup$
  – javacoder
  Mar 15 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you consider the section of the figure along the$x-z$ plane you will see it is a circular sector delimited by segments in the lines $z=x$ and $z=-x$ and an arc connecting them. So, $z$, for a fixed $x=r$, runs from that value of $x=r$ to that arc $sqrt1-r^2$
  $endgroup$
  – Rafa Budría
  Mar 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I can see it geometrically, is there way to tell that its true algebraically?
  $endgroup$
  – javacoder
  Mar 15 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, we know the limits because they come directly from the definition of the region: $S=zgeq rland zleqsqrt1-r^2$
  $endgroup$
  – Rafa Budría
  Mar 15 at 20:45
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

The limits for $theta$ are correct. For $r$, as it has to be positive, are $0leq rleqdfrac1sqrt2$. Now, the region of integration is an inverted cone with its vertex at the origin (the equation considers only the positive square root) and the sphere as its "ceiling", so, we have different formulas for the lower limit and for the upper one: $rleq zleq sqrt1-r^2$

$$V(S)=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2rmathbb dz,mathbb dr,mathbb dtheta$$

For the centroid we have to evaluate three integrals, one for each coordinate $C=(C_r,C_theta,C_z)$ but we can take advantage of the region's symmetry of rotation around the $z$ axis. The centroid is in some point of this axis $C=(0,0,C_z)$, with,

$$C_z=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2zrmathbb dz,mathbb dr,mathbb dtheta$$

share|cite|improve this answer

edited Mar 15 at 19:21

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can't understand why the lower limit for $z$ is $r$... How do you know? I understand the lower and upper limit for $r$.
  $endgroup$
  – javacoder
  Mar 15 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you consider the section of the figure along the$x-z$ plane you will see it is a circular sector delimited by segments in the lines $z=x$ and $z=-x$ and an arc connecting them. So, $z$, for a fixed $x=r$, runs from that value of $x=r$ to that arc $sqrt1-r^2$
  $endgroup$
  – Rafa Budría
  Mar 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I can see it geometrically, is there way to tell that its true algebraically?
  $endgroup$
  – javacoder
  Mar 15 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, we know the limits because they come directly from the definition of the region: $S=zgeq rland zleqsqrt1-r^2$
  $endgroup$
  – Rafa Budría
  Mar 15 at 20:45
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The limits for $theta$ are correct. For $r$, as it has to be positive, are $0leq rleqdfrac1sqrt2$. Now, the region of integration is an inverted cone with its vertex at the origin (the equation considers only the positive square root) and the sphere as its "ceiling", so, we have different formulas for the lower limit and for the upper one: $rleq zleq sqrt1-r^2$

$$V(S)=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2rmathbb dz,mathbb dr,mathbb dtheta$$

For the centroid we have to evaluate three integrals, one for each coordinate $C=(C_r,C_theta,C_z)$ but we can take advantage of the region's symmetry of rotation around the $z$ axis. The centroid is in some point of this axis $C=(0,0,C_z)$, with,

$$C_z=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2zrmathbb dz,mathbb dr,mathbb dtheta$$

share|cite|improve this answer

edited Mar 15 at 19:21

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can't understand why the lower limit for $z$ is $r$... How do you know? I understand the lower and upper limit for $r$.
  $endgroup$
  – javacoder
  Mar 15 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you consider the section of the figure along the$x-z$ plane you will see it is a circular sector delimited by segments in the lines $z=x$ and $z=-x$ and an arc connecting them. So, $z$, for a fixed $x=r$, runs from that value of $x=r$ to that arc $sqrt1-r^2$
  $endgroup$
  – Rafa Budría
  Mar 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I can see it geometrically, is there way to tell that its true algebraically?
  $endgroup$
  – javacoder
  Mar 15 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, we know the limits because they come directly from the definition of the region: $S=zgeq rland zleqsqrt1-r^2$
  $endgroup$
  – Rafa Budría
  Mar 15 at 20:45
  
  

  
  
  
  

add a comment | 

$begingroup$

The limits for $theta$ are correct. For $r$, as it has to be positive, are $0leq rleqdfrac1sqrt2$. Now, the region of integration is an inverted cone with its vertex at the origin (the equation considers only the positive square root) and the sphere as its "ceiling", so, we have different formulas for the lower limit and for the upper one: $rleq zleq sqrt1-r^2$

$$V(S)=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2rmathbb dz,mathbb dr,mathbb dtheta$$

For the centroid we have to evaluate three integrals, one for each coordinate $C=(C_r,C_theta,C_z)$ but we can take advantage of the region's symmetry of rotation around the $z$ axis. The centroid is in some point of this axis $C=(0,0,C_z)$, with,

$$C_z=int_0^2piint_0^1/sqrt2int_r^sqrt1-r^2zrmathbb dz,mathbb dr,mathbb dtheta$$

share|cite|improve this answer

edited Mar 15 at 19:21

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825

$endgroup$

share|cite|improve this answer

edited Mar 15 at 19:21

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825

answered Mar 15 at 9:43

Rafa BudríaRafa Budría

5,9101825