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What's The Cylindrical coordinates for this $int_0^6int_-sqrt6x-x^2^sqrt6x-x^2int_0^6x-x^2-y^2left(x^2+y^2right)dzdydx$

convert triple integral $int_-6^6int_-sqrt36-x^2^sqrt36-x^2int_x^2+y^2^36x,dz,dy,dx$ to spherical coordinatesConvert the integral from rectangular to cylindrical coordinates and solveFind volume of the region bounded by… using cylindrical coordinates.Problem with cartesian to cylindrical coordinatesConvert this integral to cylindrical and spherical coordinates: $int_-2^2 int_-sqrt4-x^2^sqrt4-x^2int_x^2+y^2^4 x dz dy dx$Converting a double Integral from Cartesian to Cylindrical CoordinatesConverting the coordinates of a point to cylindrical coordinates with positive values.How to evaluate the integrals in the cylindrical coordinatesCalculate this triple integral in cylindrical coordinates, the result is different with triple integral in cartesian coordinatesUsing cylindrical coordinates, find the volume of the region $D =y^2+z^2le5+x^2,4x^2+y^2+z^2le25$

-1

$begingroup$

I want to convert this to cylindrical coordinates
$$V=int_0^6int_-sqrt6x-x^2^sqrt6x-x^2int_0^6x-x^2-y^2left(x^2+y^2right)dzdydx = 486π$$
I want to write it like this:
$$V=int_ ^ int_ ^ int_ ^ r^3dzdrdtheta$$
if you don't know how to do it at least vote please!

calculus integration definite-integrals cylindrical-coordinates

share|cite|improve this question

edited Mar 22 at 15:56

jon wick

asked Mar 22 at 15:28

jon wickjon wick

157

$endgroup$

add a comment | 

-1

$begingroup$

I want to convert this to cylindrical coordinates
$$V=int_0^6int_-sqrt6x-x^2^sqrt6x-x^2int_0^6x-x^2-y^2left(x^2+y^2right)dzdydx = 486π$$
I want to write it like this:
$$V=int_ ^ int_ ^ int_ ^ r^3dzdrdtheta$$
if you don't know how to do it at least vote please!

calculus integration definite-integrals cylindrical-coordinates

share|cite|improve this question

edited Mar 22 at 15:56

jon wick

asked Mar 22 at 15:28

jon wickjon wick

157

$endgroup$

add a comment | 

$begingroup$

I want to convert this to cylindrical coordinates
$$V=int_0^6int_-sqrt6x-x^2^sqrt6x-x^2int_0^6x-x^2-y^2left(x^2+y^2right)dzdydx = 486π$$
I want to write it like this:
$$V=int_ ^ int_ ^ int_ ^ r^3dzdrdtheta$$
if you don't know how to do it at least vote please!

calculus integration definite-integrals cylindrical-coordinates

share|cite|improve this question

edited Mar 22 at 15:56

jon wick

asked Mar 22 at 15:28

jon wickjon wick

157

$endgroup$

share|cite|improve this question

edited Mar 22 at 15:56

jon wick

asked Mar 22 at 15:28

jon wickjon wick

157

share|cite|improve this question

edited Mar 22 at 15:56

jon wick

asked Mar 22 at 15:28

jon wickjon wick

157

asked Mar 22 at 15:28

jon wickjon wick

157

asked Mar 22 at 15:28

jon wickjon wick

157

2 Answers
2

active

oldest

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0

$begingroup$

$zin[0,6x-x^2-y^2]$ tells you the region is "vertically" bounded between the plane $z=0$ and the paraboloid $z=6x-x^2-y^2$. Complete the square to write the latter as $z=9-(x-3)^2-y^2$, which is a paraboloid whose vertex is located at $(3,0,9)$.

$yin[-sqrt6x-x^2,sqrt6x-x^2]$, or $yin[-sqrt9-(x-3)^2,sqrt9-(x-3)^2]$, tells you that for any $z$, $y$ is confined by the upper and lower halves of a circle with radius $3$ and centered at $(3,0,z)$. This describes a cylinder in $mathbb R^3$.

$xin[0,6]$ tells you the integration region covers the interior of the cylinder.

Now you can convert to a modified cylindrical coordinate system with

$$begincasesx=rcostheta+3\y=rsintheta\z=zendcases$$

and the region you're interested in (I'll call it $E$) is obtained with $0le rle3$, $0lethetale2pi$, and $0le zle9-r^2$. The Jacobian is the same as with the conversion to standard cylindrical coordinates. So the integral is

$$beginalign*
iiint_E(x^2+y^2),mathrm dV&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2left((rcostheta+3)^2+(rsintheta)^2right)r,mathrm dz,mathrm dr,mathrm dtheta\[1ex]
&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2(r^2+6rcostheta+9)r,mathrm dz,mathrm dr,mathrm dtheta
endalign*$$

share|cite|improve this answer

answered Mar 22 at 16:45

user170231user170231

4,21511429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks very much that's Correct
  $endgroup$
  – jon wick
  Mar 22 at 17:38
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: the limits give the set:
$$0le xle 6,$$
$$y^2le 6x - x^2,$$
$$0le zle 6x - x^2 - y^2.$$
Now, do $x = rcostheta$, $y = rsintheta$...

share|cite|improve this answer

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

add a comment | 

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0

$begingroup$

$zin[0,6x-x^2-y^2]$ tells you the region is "vertically" bounded between the plane $z=0$ and the paraboloid $z=6x-x^2-y^2$. Complete the square to write the latter as $z=9-(x-3)^2-y^2$, which is a paraboloid whose vertex is located at $(3,0,9)$.

$yin[-sqrt6x-x^2,sqrt6x-x^2]$, or $yin[-sqrt9-(x-3)^2,sqrt9-(x-3)^2]$, tells you that for any $z$, $y$ is confined by the upper and lower halves of a circle with radius $3$ and centered at $(3,0,z)$. This describes a cylinder in $mathbb R^3$.

$xin[0,6]$ tells you the integration region covers the interior of the cylinder.

Now you can convert to a modified cylindrical coordinate system with

$$begincasesx=rcostheta+3\y=rsintheta\z=zendcases$$

and the region you're interested in (I'll call it $E$) is obtained with $0le rle3$, $0lethetale2pi$, and $0le zle9-r^2$. The Jacobian is the same as with the conversion to standard cylindrical coordinates. So the integral is

$$beginalign*
iiint_E(x^2+y^2),mathrm dV&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2left((rcostheta+3)^2+(rsintheta)^2right)r,mathrm dz,mathrm dr,mathrm dtheta\[1ex]
&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2(r^2+6rcostheta+9)r,mathrm dz,mathrm dr,mathrm dtheta
endalign*$$

share|cite|improve this answer

answered Mar 22 at 16:45

user170231user170231

4,21511429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks very much that's Correct
  $endgroup$
  – jon wick
  Mar 22 at 17:38
  

  
  
  
  

add a comment | 

0

$begingroup$

$zin[0,6x-x^2-y^2]$ tells you the region is "vertically" bounded between the plane $z=0$ and the paraboloid $z=6x-x^2-y^2$. Complete the square to write the latter as $z=9-(x-3)^2-y^2$, which is a paraboloid whose vertex is located at $(3,0,9)$.

$yin[-sqrt6x-x^2,sqrt6x-x^2]$, or $yin[-sqrt9-(x-3)^2,sqrt9-(x-3)^2]$, tells you that for any $z$, $y$ is confined by the upper and lower halves of a circle with radius $3$ and centered at $(3,0,z)$. This describes a cylinder in $mathbb R^3$.

$xin[0,6]$ tells you the integration region covers the interior of the cylinder.

Now you can convert to a modified cylindrical coordinate system with

$$begincasesx=rcostheta+3\y=rsintheta\z=zendcases$$

and the region you're interested in (I'll call it $E$) is obtained with $0le rle3$, $0lethetale2pi$, and $0le zle9-r^2$. The Jacobian is the same as with the conversion to standard cylindrical coordinates. So the integral is

$$beginalign*
iiint_E(x^2+y^2),mathrm dV&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2left((rcostheta+3)^2+(rsintheta)^2right)r,mathrm dz,mathrm dr,mathrm dtheta\[1ex]
&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2(r^2+6rcostheta+9)r,mathrm dz,mathrm dr,mathrm dtheta
endalign*$$

share|cite|improve this answer

answered Mar 22 at 16:45

user170231user170231

4,21511429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks very much that's Correct
  $endgroup$
  – jon wick
  Mar 22 at 17:38
  

  
  
  
  

add a comment | 

$begingroup$

$zin[0,6x-x^2-y^2]$ tells you the region is "vertically" bounded between the plane $z=0$ and the paraboloid $z=6x-x^2-y^2$. Complete the square to write the latter as $z=9-(x-3)^2-y^2$, which is a paraboloid whose vertex is located at $(3,0,9)$.

$yin[-sqrt6x-x^2,sqrt6x-x^2]$, or $yin[-sqrt9-(x-3)^2,sqrt9-(x-3)^2]$, tells you that for any $z$, $y$ is confined by the upper and lower halves of a circle with radius $3$ and centered at $(3,0,z)$. This describes a cylinder in $mathbb R^3$.

$xin[0,6]$ tells you the integration region covers the interior of the cylinder.

Now you can convert to a modified cylindrical coordinate system with

$$begincasesx=rcostheta+3\y=rsintheta\z=zendcases$$

and the region you're interested in (I'll call it $E$) is obtained with $0le rle3$, $0lethetale2pi$, and $0le zle9-r^2$. The Jacobian is the same as with the conversion to standard cylindrical coordinates. So the integral is

$$beginalign*
iiint_E(x^2+y^2),mathrm dV&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2left((rcostheta+3)^2+(rsintheta)^2right)r,mathrm dz,mathrm dr,mathrm dtheta\[1ex]
&=int_theta=0^theta=2piint_r=0^r=3int_z=0^z=9-r^2(r^2+6rcostheta+9)r,mathrm dz,mathrm dr,mathrm dtheta
endalign*$$

share|cite|improve this answer

answered Mar 22 at 16:45

user170231user170231

4,21511429

$endgroup$

share|cite|improve this answer

answered Mar 22 at 16:45

user170231user170231

4,21511429

answered Mar 22 at 16:45

user170231user170231

4,21511429

answered Mar 22 at 16:45

user170231user170231

4,21511429

0

$begingroup$

Hint: the limits give the set:
$$0le xle 6,$$
$$y^2le 6x - x^2,$$
$$0le zle 6x - x^2 - y^2.$$
Now, do $x = rcostheta$, $y = rsintheta$...

share|cite|improve this answer

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

add a comment | 

0

$begingroup$

Hint: the limits give the set:
$$0le xle 6,$$
$$y^2le 6x - x^2,$$
$$0le zle 6x - x^2 - y^2.$$
Now, do $x = rcostheta$, $y = rsintheta$...

share|cite|improve this answer

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

add a comment | 

$begingroup$

Hint: the limits give the set:
$$0le xle 6,$$
$$y^2le 6x - x^2,$$
$$0le zle 6x - x^2 - y^2.$$
Now, do $x = rcostheta$, $y = rsintheta$...

share|cite|improve this answer

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

share|cite|improve this answer

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

answered Mar 22 at 16:45

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972