Mixing
Spherical coordinates in surface integrals
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am stuck on the following problem
Evaluate :
$$I=iint x^2 y^2 z dS $$
where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= frac 2pi a^7 105$$
surface-integrals
asked Jul 16 at 14:41
Speedy
65
add a comment |Â
up vote
1
down vote
favorite
I am stuck on the following problem
Evaluate :
$$I=iint x^2 y^2 z dS $$
where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= frac 2pi a^7 105$$
surface-integrals
asked Jul 16 at 14:41
Speedy
65
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 14:44
1
What means ... positive side of lower half of the sphere?
– Cesareo
Jul 16 at 15:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am stuck on the following problem
Evaluate :
$$I=iint x^2 y^2 z dS $$
where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= frac 2pi a^7 105$$
surface-integrals
asked Jul 16 at 14:41
Speedy
65
I am stuck on the following problem
Evaluate :
$$I=iint x^2 y^2 z dS $$
where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= frac 2pi a^7 105$$
surface-integrals
asked Jul 16 at 14:41
Speedy
65
edited Jul 16 at 14:55
asked Jul 16 at 14:41
Speedy
65
asked Jul 16 at 14:41
Speedy
65
asked Jul 16 at 14:41
Speedy
65
65
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 14:44
1
What means ... positive side of lower half of the sphere?
– Cesareo
Jul 16 at 15:23
add a comment |Â
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 14:44
1
What means ... positive side of lower half of the sphere?
– Cesareo
Jul 16 at 15:23
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 14:44
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 14:44
1
1
What means ... positive side of lower half of the sphere?
– Cesareo
Jul 16 at 15:23
What means ... positive side of lower half of the sphere?
– Cesareo
Jul 16 at 15:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Consider the following transformation:
beginequation
beginsplit
x&=asinphicostheta\
y&=asinphisintheta\
z&=acosphi
endsplit
endequation
where $thetain[0,2pi]$ and $phiin[-fracpi2,0]$, then $dS=a^2sinphi;dphi;dtheta$. Now
$$
iint_Sx^2y^2z:dS=a^7int_0^2picos^2thetasin^2theta;dthetaint_-pi/2^0sin^5phicosphi;dphi=a^7bigg[fracpi4bigg]cdotbigg[-frac16bigg]=-fracpi a^724.
$$
Note that
$$int_-pi/2^0sin^5phicosphi;dphi=int_-pi/2^0sin^5phi(sinphi)';dphi=bigg[frac16sin^6phibigg]_-pi/2^0=-frac16
$$
answered Jul 16 at 16:28
Jack J.
3661317
add a comment |Â
up vote
0
down vote
Substitute $z=-sqrta^2-x^2-y^2$ and $dS=sqrt1+z_x^2+z_y^2dA$. which turn $$I=-aiint_R x^2y^2dA.$$ Use polar co-ordinate.$x=rcostheta, y=rsin theta$. $dxdy=rdrdtheta$. $r$ varies from $0$ to $a$ and $theta$ varies from $0$ to $2pi$. Which turn $I=-aint_0^2pisin^2theta cos^2theta dtheta int_0^a r^5 dr=frac-a^76int_0^2pisin^2theta cos^2theta dtheta =frac-a^724int_0^2pisin^2 2theta dtheta=frac-a^748int_0^2pi(1-cos4theta) dtheta=frac-a^7pi24$
answered Jul 16 at 14:57
N. Maneesh
2,4271924
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Consider the following transformation:
beginequation
beginsplit
x&=asinphicostheta\
y&=asinphisintheta\
z&=acosphi
endsplit
endequation
where $thetain[0,2pi]$ and $phiin[-fracpi2,0]$, then $dS=a^2sinphi;dphi;dtheta$. Now
$$
iint_Sx^2y^2z:dS=a^7int_0^2picos^2thetasin^2theta;dthetaint_-pi/2^0sin^5phicosphi;dphi=a^7bigg[fracpi4bigg]cdotbigg[-frac16bigg]=-fracpi a^724.
$$
Note that
$$int_-pi/2^0sin^5phicosphi;dphi=int_-pi/2^0sin^5phi(sinphi)';dphi=bigg[frac16sin^6phibigg]_-pi/2^0=-frac16
$$
answered Jul 16 at 16:28
Jack J.
3661317
add a comment |Â
up vote
0
down vote
accepted
Consider the following transformation:
beginequation
beginsplit
x&=asinphicostheta\
y&=asinphisintheta\
z&=acosphi
endsplit
endequation
where $thetain[0,2pi]$ and $phiin[-fracpi2,0]$, then $dS=a^2sinphi;dphi;dtheta$. Now
$$
iint_Sx^2y^2z:dS=a^7int_0^2picos^2thetasin^2theta;dthetaint_-pi/2^0sin^5phicosphi;dphi=a^7bigg[fracpi4bigg]cdotbigg[-frac16bigg]=-fracpi a^724.
$$
Note that
$$int_-pi/2^0sin^5phicosphi;dphi=int_-pi/2^0sin^5phi(sinphi)';dphi=bigg[frac16sin^6phibigg]_-pi/2^0=-frac16
$$
answered Jul 16 at 16:28
Jack J.
3661317
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Consider the following transformation:
beginequation
beginsplit
x&=asinphicostheta\
y&=asinphisintheta\
z&=acosphi
endsplit
endequation
where $thetain[0,2pi]$ and $phiin[-fracpi2,0]$, then $dS=a^2sinphi;dphi;dtheta$. Now
$$
iint_Sx^2y^2z:dS=a^7int_0^2picos^2thetasin^2theta;dthetaint_-pi/2^0sin^5phicosphi;dphi=a^7bigg[fracpi4bigg]cdotbigg[-frac16bigg]=-fracpi a^724.
$$
Note that
$$int_-pi/2^0sin^5phicosphi;dphi=int_-pi/2^0sin^5phi(sinphi)';dphi=bigg[frac16sin^6phibigg]_-pi/2^0=-frac16
$$
answered Jul 16 at 16:28
Jack J.
3661317
Consider the following transformation:
beginequation
beginsplit
x&=asinphicostheta\
y&=asinphisintheta\
z&=acosphi
endsplit
endequation
where $thetain[0,2pi]$ and $phiin[-fracpi2,0]$, then $dS=a^2sinphi;dphi;dtheta$. Now
$$
iint_Sx^2y^2z:dS=a^7int_0^2picos^2thetasin^2theta;dthetaint_-pi/2^0sin^5phicosphi;dphi=a^7bigg[fracpi4bigg]cdotbigg[-frac16bigg]=-fracpi a^724.
$$
Note that
$$int_-pi/2^0sin^5phicosphi;dphi=int_-pi/2^0sin^5phi(sinphi)';dphi=bigg[frac16sin^6phibigg]_-pi/2^0=-frac16
$$
answered Jul 16 at 16:28
Jack J.
3661317
edited Jul 16 at 16:38
answered Jul 16 at 16:28
Jack J.
3661317
answered Jul 16 at 16:28
Jack J.
3661317
answered Jul 16 at 16:28
Jack J.
3661317
3661317
add a comment |Â
add a comment |Â
up vote
0
down vote
Substitute $z=-sqrta^2-x^2-y^2$ and $dS=sqrt1+z_x^2+z_y^2dA$. which turn $$I=-aiint_R x^2y^2dA.$$ Use polar co-ordinate.$x=rcostheta, y=rsin theta$. $dxdy=rdrdtheta$. $r$ varies from $0$ to $a$ and $theta$ varies from $0$ to $2pi$. Which turn $I=-aint_0^2pisin^2theta cos^2theta dtheta int_0^a r^5 dr=frac-a^76int_0^2pisin^2theta cos^2theta dtheta =frac-a^724int_0^2pisin^2 2theta dtheta=frac-a^748int_0^2pi(1-cos4theta) dtheta=frac-a^7pi24$
answered Jul 16 at 14:57
N. Maneesh
2,4271924
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
 |Â
show 4 more comments
up vote
0
down vote
Substitute $z=-sqrta^2-x^2-y^2$ and $dS=sqrt1+z_x^2+z_y^2dA$. which turn $$I=-aiint_R x^2y^2dA.$$ Use polar co-ordinate.$x=rcostheta, y=rsin theta$. $dxdy=rdrdtheta$. $r$ varies from $0$ to $a$ and $theta$ varies from $0$ to $2pi$. Which turn $I=-aint_0^2pisin^2theta cos^2theta dtheta int_0^a r^5 dr=frac-a^76int_0^2pisin^2theta cos^2theta dtheta =frac-a^724int_0^2pisin^2 2theta dtheta=frac-a^748int_0^2pi(1-cos4theta) dtheta=frac-a^7pi24$
answered Jul 16 at 14:57
N. Maneesh
2,4271924
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
 |Â
show 4 more comments
up vote
0
down vote
up vote
0
down vote
Substitute $z=-sqrta^2-x^2-y^2$ and $dS=sqrt1+z_x^2+z_y^2dA$. which turn $$I=-aiint_R x^2y^2dA.$$ Use polar co-ordinate.$x=rcostheta, y=rsin theta$. $dxdy=rdrdtheta$. $r$ varies from $0$ to $a$ and $theta$ varies from $0$ to $2pi$. Which turn $I=-aint_0^2pisin^2theta cos^2theta dtheta int_0^a r^5 dr=frac-a^76int_0^2pisin^2theta cos^2theta dtheta =frac-a^724int_0^2pisin^2 2theta dtheta=frac-a^748int_0^2pi(1-cos4theta) dtheta=frac-a^7pi24$
answered Jul 16 at 14:57
N. Maneesh
2,4271924
Substitute $z=-sqrta^2-x^2-y^2$ and $dS=sqrt1+z_x^2+z_y^2dA$. which turn $$I=-aiint_R x^2y^2dA.$$ Use polar co-ordinate.$x=rcostheta, y=rsin theta$. $dxdy=rdrdtheta$. $r$ varies from $0$ to $a$ and $theta$ varies from $0$ to $2pi$. Which turn $I=-aint_0^2pisin^2theta cos^2theta dtheta int_0^a r^5 dr=frac-a^76int_0^2pisin^2theta cos^2theta dtheta =frac-a^724int_0^2pisin^2 2theta dtheta=frac-a^748int_0^2pi(1-cos4theta) dtheta=frac-a^7pi24$
answered Jul 16 at 14:57
N. Maneesh
2,4271924
edited Jul 16 at 15:23
answered Jul 16 at 14:57
N. Maneesh
2,4271924
answered Jul 16 at 14:57
N. Maneesh
2,4271924
answered Jul 16 at 14:57
N. Maneesh
2,4271924
2,4271924
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
 |Â
show 4 more comments
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
I tried this method also ....... I am getting the same answer( a^2 )pi / 24
– Speedy
Jul 16 at 15:10
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
your answer must be wrong. power of $a$ must be $7$. It is easy to see.
– N. Maneesh
Jul 16 at 15:13
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Oh sorry my typo error it is (a^7) pi / 24
– Speedy
Jul 16 at 15:14
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
Textbook answer is wrong.
– N. Maneesh
Jul 16 at 15:28
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
There are other questions also which are not reaching to their correct conclusion..... Should i post them too ?
– Speedy
Jul 16 at 16:08
 |Â
show 4 more comments
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853473%2fspherical-coordinates-in-surface-integrals%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 16 at 14:44
1
What means ... positive side of lower half of the sphere?
– Cesareo
Jul 16 at 15:23