Integration over ellipsoid using the divergence theorem

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I have troubles with the following question:

Given is the following vector field :
$$f(x,y,z)= beginpmatrixx+sin(y)cos(z)\y+e^x^2+y^2\-z + log(1+x^2+y^2)endpmatrix$$

and the set $M$ is defined by
$$M:=(x,y,z)mid x^2+fracy^24+fracz^29=1$$

Calculate the following integral :
$$int_Mlangle f, v rangle, dS$$
where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.

---

So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^x^2+y^2$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_textEllipsoid 2ye^x^2+y^2 dx,dy,dz$ ?

Thanks for your help.

real-analysis integration

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edited Jul 18 at 19:37

Robert Howard

1,327620

asked Jul 18 at 19:18

Poujh

182112

add a comment | 

up vote
0
down vote

favorite

I have troubles with the following question:

Given is the following vector field :
$$f(x,y,z)= beginpmatrixx+sin(y)cos(z)\y+e^x^2+y^2\-z + log(1+x^2+y^2)endpmatrix$$

and the set $M$ is defined by
$$M:=(x,y,z)mid x^2+fracy^24+fracz^29=1$$

Calculate the following integral :
$$int_Mlangle f, v rangle, dS$$
where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.

---

So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^x^2+y^2$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_textEllipsoid 2ye^x^2+y^2 dx,dy,dz$ ?

Thanks for your help.

real-analysis integration

share|cite|improve this question

edited Jul 18 at 19:37

Robert Howard

1,327620

asked Jul 18 at 19:18

Poujh

182112

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I have troubles with the following question:

Given is the following vector field :
$$f(x,y,z)= beginpmatrixx+sin(y)cos(z)\y+e^x^2+y^2\-z + log(1+x^2+y^2)endpmatrix$$

and the set $M$ is defined by
$$M:=(x,y,z)mid x^2+fracy^24+fracz^29=1$$

Calculate the following integral :
$$int_Mlangle f, v rangle, dS$$
where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.

---

So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^x^2+y^2$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_textEllipsoid 2ye^x^2+y^2 dx,dy,dz$ ?

Thanks for your help.

real-analysis integration

share|cite|improve this question

edited Jul 18 at 19:37

Robert Howard

1,327620

asked Jul 18 at 19:18

Poujh

182112

I have troubles with the following question:

Given is the following vector field :
$$f(x,y,z)= beginpmatrixx+sin(y)cos(z)\y+e^x^2+y^2\-z + log(1+x^2+y^2)endpmatrix$$

and the set $M$ is defined by
$$M:=(x,y,z)mid x^2+fracy^24+fracz^29=1$$

Calculate the following integral :
$$int_Mlangle f, v rangle, dS$$
where $v$ is the unit normal field (german : Einheitsnormalenfeld) toward the exterior.

---

So I thought about using the divergence theorem as usual given that $M$ is just the border of an ellipsoid. I get the triple integral over $1+2ye^x^2+y^2$ . The 3 integrals over 1 give us the volume of a normal ellipsoid, in this case $8pi$, but how do I integrate $iiint_textEllipsoid 2ye^x^2+y^2 dx,dy,dz$ ?

Thanks for your help.

real-analysis integration

share|cite|improve this question

edited Jul 18 at 19:37

Robert Howard

1,327620

asked Jul 18 at 19:18

Poujh

182112

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 19:37

Robert Howard

1,327620

edited Jul 18 at 19:37

Robert Howard

1,327620

edited Jul 18 at 19:37

Robert Howard

1,327620

1,327620

asked Jul 18 at 19:18

Poujh

182112

asked Jul 18 at 19:18

Poujh

182112

asked Jul 18 at 19:18

Poujh

182112

182112

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

So you are asking how to set up a triple integral?

The ellipsoid $x^2+ fracy^24+ fracz^29= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt1-x^2$ to $2sqrt1-x^2$; and $z$, for each $x$ and $y$, from $-3sqrt1-x^2-fracy^24$ to $3sqrt1-x^2-fracy^24$.

So the integral will be $$int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2int_z=-3sqrt1-x^2-fracy^24^3sqrt1-x^2-fracy^24left(1+ 2ye^x^2+y^2right)dz,dy,dx$$

Since the integrand is independent of $z$, the first integration gives
$$-6int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2sqrt1- x^2left(1+ 2ye^x^2+y^2right)dy,dx$$

share|cite|improve this answer

edited Jul 18 at 20:02

Robert Howard

1,327620

answered Jul 18 at 19:36

user247327

9,6781515

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
  – Poujh
  Jul 18 at 19:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
  – zokomoko
  Jul 18 at 20:07
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

So you are asking how to set up a triple integral?

The ellipsoid $x^2+ fracy^24+ fracz^29= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt1-x^2$ to $2sqrt1-x^2$; and $z$, for each $x$ and $y$, from $-3sqrt1-x^2-fracy^24$ to $3sqrt1-x^2-fracy^24$.

So the integral will be $$int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2int_z=-3sqrt1-x^2-fracy^24^3sqrt1-x^2-fracy^24left(1+ 2ye^x^2+y^2right)dz,dy,dx$$

Since the integrand is independent of $z$, the first integration gives
$$-6int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2sqrt1- x^2left(1+ 2ye^x^2+y^2right)dy,dx$$

share|cite|improve this answer

edited Jul 18 at 20:02

Robert Howard

1,327620

answered Jul 18 at 19:36

user247327

9,6781515

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
  – Poujh
  Jul 18 at 19:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
  – zokomoko
  Jul 18 at 20:07
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

So you are asking how to set up a triple integral?

The ellipsoid $x^2+ fracy^24+ fracz^29= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt1-x^2$ to $2sqrt1-x^2$; and $z$, for each $x$ and $y$, from $-3sqrt1-x^2-fracy^24$ to $3sqrt1-x^2-fracy^24$.

So the integral will be $$int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2int_z=-3sqrt1-x^2-fracy^24^3sqrt1-x^2-fracy^24left(1+ 2ye^x^2+y^2right)dz,dy,dx$$

Since the integrand is independent of $z$, the first integration gives
$$-6int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2sqrt1- x^2left(1+ 2ye^x^2+y^2right)dy,dx$$

share|cite|improve this answer

edited Jul 18 at 20:02

Robert Howard

1,327620

answered Jul 18 at 19:36

user247327

9,6781515

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
  – Poujh
  Jul 18 at 19:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
  – zokomoko
  Jul 18 at 20:07
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

So you are asking how to set up a triple integral?

The ellipsoid $x^2+ fracy^24+ fracz^29= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt1-x^2$ to $2sqrt1-x^2$; and $z$, for each $x$ and $y$, from $-3sqrt1-x^2-fracy^24$ to $3sqrt1-x^2-fracy^24$.

So the integral will be $$int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2int_z=-3sqrt1-x^2-fracy^24^3sqrt1-x^2-fracy^24left(1+ 2ye^x^2+y^2right)dz,dy,dx$$

Since the integrand is independent of $z$, the first integration gives
$$-6int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2sqrt1- x^2left(1+ 2ye^x^2+y^2right)dy,dx$$

share|cite|improve this answer

edited Jul 18 at 20:02

Robert Howard

1,327620

answered Jul 18 at 19:36

user247327

9,6781515

So you are asking how to set up a triple integral?

The ellipsoid $x^2+ fracy^24+ fracz^29= 1$ can be covered by taking $x$ from $-1$ to $1$; $y$, for each $x$, from $-2sqrt1-x^2$ to $2sqrt1-x^2$; and $z$, for each $x$ and $y$, from $-3sqrt1-x^2-fracy^24$ to $3sqrt1-x^2-fracy^24$.

So the integral will be $$int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2int_z=-3sqrt1-x^2-fracy^24^3sqrt1-x^2-fracy^24left(1+ 2ye^x^2+y^2right)dz,dy,dx$$

Since the integrand is independent of $z$, the first integration gives
$$-6int_x=-1^1int_y=-2sqrt1- x^2^2sqrt1- x^2sqrt1- x^2left(1+ 2ye^x^2+y^2right)dy,dx$$

share|cite|improve this answer

edited Jul 18 at 20:02

Robert Howard

1,327620

answered Jul 18 at 19:36

user247327

9,6781515

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 20:02

Robert Howard

1,327620

edited Jul 18 at 20:02

Robert Howard

1,327620

edited Jul 18 at 20:02

Robert Howard

1,327620

1,327620

answered Jul 18 at 19:36

user247327

9,6781515

answered Jul 18 at 19:36

user247327

9,6781515

answered Jul 18 at 19:36

user247327

9,6781515

9,6781515

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
  – Poujh
  Jul 18 at 19:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
  – zokomoko
  Jul 18 at 20:07
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
  – Poujh
  Jul 18 at 19:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
  – zokomoko
  Jul 18 at 20:07
  
  

  
  
  
  

1

1

Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
– Poujh
Jul 18 at 19:42

Thanks for your help, but isn't that really "complicated" ? I mean, the integral over 1 is easy, and the divergence of the vector field is also a lot easier than in "first" appearence, so do we really need to integrate $2ye^x^2+y^2$ in that way ? Isn't there a trick, or couldn't we for example use cylindrical/ spherical coordinates to make things easier ? I found this question in an old exam, and given that time in exam is often limited, the questions are generally not "that" hard, so I'm just a bit surprised.
– Poujh
Jul 18 at 19:42

The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
– zokomoko
Jul 18 at 20:07

The first integral is an integral of an odd function over a symmetric interval, thus the entire calculation yields zero.
– zokomoko
Jul 18 at 20:07

add a comment | 

 

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