Mixing
Evaluating Double Integral
$begingroup$
Evaluate $$iint_{S} (x^3+y^3),dy,dz + (z^3+y^3),dz,dx + (x^3+z^3),dx,dy$$ where $S$ is the surface of the sphere $x^2+y^2+z^2= a^2$.
Spherical coordinates help me in this question, $r^2=a^2$.
We can also write $$iiint_{S} 2(x^3+y^3+z^3),dx,dy,dz$$ or not?
Thank you for your advanced ideas.
calculus integration indefinite-integrals
edited Mar 19 at 23:41
Robert Howard
2,3033935
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
$endgroup$
add a comment |
$begingroup$
Evaluate $$iint_{S} (x^3+y^3),dy,dz + (z^3+y^3),dz,dx + (x^3+z^3),dx,dy$$ where $S$ is the surface of the sphere $x^2+y^2+z^2= a^2$.
Spherical coordinates help me in this question, $r^2=a^2$.
We can also write $$iiint_{S} 2(x^3+y^3+z^3),dx,dy,dz$$ or not?
Thank you for your advanced ideas.
calculus integration indefinite-integrals
edited Mar 19 at 23:41
Robert Howard
2,3033935
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
$endgroup$
add a comment |
$begingroup$
Evaluate $$iint_{S} (x^3+y^3),dy,dz + (z^3+y^3),dz,dx + (x^3+z^3),dx,dy$$ where $S$ is the surface of the sphere $x^2+y^2+z^2= a^2$.
Spherical coordinates help me in this question, $r^2=a^2$.
We can also write $$iiint_{S} 2(x^3+y^3+z^3),dx,dy,dz$$ or not?
Thank you for your advanced ideas.
calculus integration indefinite-integrals
edited Mar 19 at 23:41
Robert Howard
2,3033935
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
$endgroup$
Evaluate $$iint_{S} (x^3+y^3),dy,dz + (z^3+y^3),dz,dx + (x^3+z^3),dx,dy$$ where $S$ is the surface of the sphere $x^2+y^2+z^2= a^2$.
Spherical coordinates help me in this question, $r^2=a^2$.
We can also write $$iiint_{S} 2(x^3+y^3+z^3),dx,dy,dz$$ or not?
Thank you for your advanced ideas.
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Mar 19 at 23:41
Robert Howard
2,3033935
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
edited Mar 19 at 23:41
Robert Howard
2,3033935
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
edited Mar 19 at 23:41
Robert Howard
2,3033935
edited Mar 19 at 23:41
Robert Howard
2,3033935
edited Mar 19 at 23:41
Robert Howard
2,3033935
2,3033935
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
asked Feb 1 at 8:58
mathsstudentmathsstudent
536
536
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Based on Mikhail Ostrogradsky a surface integral can be converted to volume integral where P, Q, R are function of x, y, z.
$$iint_SPmathop{dy}mathop{dz} + Qmathop{dz}mathop{dx} + Rmathop{dx}mathop{dy} = iiint_V(frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z})mathop{dx}mathop{dy}mathop{dz}$$
Hence the surface integral becomes
$$iiint_V(3x^2 + 3y^2 + 3z^2)dxmathop{dy}mathop{dz}$$
$$= 3iiint_V r^2mathop{dx}mathop{dy}mathop{dz}$$
$$= 3int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}}int_{0}^{2pi} r^4cosphi mathop{dtheta}mathop{dphi}mathop{dr}$$
$$= 6pi int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}} r^4cosphi mathop{dphi}mathop{dr}$$
$$= 12pi int_{0}^{a} r^4 mathop{dr}$$
$$= frac{12}{5}pi a^5$$
edited Mar 19 at 22:22
zwim
12.7k832
answered Mar 19 at 20:42
KY TangKY Tang
50436
$endgroup$
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
Note you can useiint_S$iint_S$ andiiint_V$iiint_V$ for multidimensional integrals, it renders better than succession ofintint$intint$
$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
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
$begingroup$
Based on Mikhail Ostrogradsky a surface integral can be converted to volume integral where P, Q, R are function of x, y, z.
$$iint_SPmathop{dy}mathop{dz} + Qmathop{dz}mathop{dx} + Rmathop{dx}mathop{dy} = iiint_V(frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z})mathop{dx}mathop{dy}mathop{dz}$$
Hence the surface integral becomes
$$iiint_V(3x^2 + 3y^2 + 3z^2)dxmathop{dy}mathop{dz}$$
$$= 3iiint_V r^2mathop{dx}mathop{dy}mathop{dz}$$
$$= 3int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}}int_{0}^{2pi} r^4cosphi mathop{dtheta}mathop{dphi}mathop{dr}$$
$$= 6pi int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}} r^4cosphi mathop{dphi}mathop{dr}$$
$$= 12pi int_{0}^{a} r^4 mathop{dr}$$
$$= frac{12}{5}pi a^5$$
edited Mar 19 at 22:22
zwim
12.7k832
answered Mar 19 at 20:42
KY TangKY Tang
50436
$endgroup$
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
Note you can useiint_S$iint_S$ andiiint_V$iiint_V$ for multidimensional integrals, it renders better than succession ofintint$intint$
$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
add a comment |
$begingroup$
Based on Mikhail Ostrogradsky a surface integral can be converted to volume integral where P, Q, R are function of x, y, z.
$$iint_SPmathop{dy}mathop{dz} + Qmathop{dz}mathop{dx} + Rmathop{dx}mathop{dy} = iiint_V(frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z})mathop{dx}mathop{dy}mathop{dz}$$
Hence the surface integral becomes
$$iiint_V(3x^2 + 3y^2 + 3z^2)dxmathop{dy}mathop{dz}$$
$$= 3iiint_V r^2mathop{dx}mathop{dy}mathop{dz}$$
$$= 3int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}}int_{0}^{2pi} r^4cosphi mathop{dtheta}mathop{dphi}mathop{dr}$$
$$= 6pi int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}} r^4cosphi mathop{dphi}mathop{dr}$$
$$= 12pi int_{0}^{a} r^4 mathop{dr}$$
$$= frac{12}{5}pi a^5$$
edited Mar 19 at 22:22
zwim
12.7k832
answered Mar 19 at 20:42
KY TangKY Tang
50436
$endgroup$
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
Note you can useiint_S$iint_S$ andiiint_V$iiint_V$ for multidimensional integrals, it renders better than succession ofintint$intint$
$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
add a comment |
$begingroup$
Based on Mikhail Ostrogradsky a surface integral can be converted to volume integral where P, Q, R are function of x, y, z.
$$iint_SPmathop{dy}mathop{dz} + Qmathop{dz}mathop{dx} + Rmathop{dx}mathop{dy} = iiint_V(frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z})mathop{dx}mathop{dy}mathop{dz}$$
Hence the surface integral becomes
$$iiint_V(3x^2 + 3y^2 + 3z^2)dxmathop{dy}mathop{dz}$$
$$= 3iiint_V r^2mathop{dx}mathop{dy}mathop{dz}$$
$$= 3int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}}int_{0}^{2pi} r^4cosphi mathop{dtheta}mathop{dphi}mathop{dr}$$
$$= 6pi int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}} r^4cosphi mathop{dphi}mathop{dr}$$
$$= 12pi int_{0}^{a} r^4 mathop{dr}$$
$$= frac{12}{5}pi a^5$$
edited Mar 19 at 22:22
zwim
12.7k832
answered Mar 19 at 20:42
KY TangKY Tang
50436
$endgroup$
Based on Mikhail Ostrogradsky a surface integral can be converted to volume integral where P, Q, R are function of x, y, z.
$$iint_SPmathop{dy}mathop{dz} + Qmathop{dz}mathop{dx} + Rmathop{dx}mathop{dy} = iiint_V(frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z})mathop{dx}mathop{dy}mathop{dz}$$
Hence the surface integral becomes
$$iiint_V(3x^2 + 3y^2 + 3z^2)dxmathop{dy}mathop{dz}$$
$$= 3iiint_V r^2mathop{dx}mathop{dy}mathop{dz}$$
$$= 3int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}}int_{0}^{2pi} r^4cosphi mathop{dtheta}mathop{dphi}mathop{dr}$$
$$= 6pi int_{0}^{a}int_{-frac{pi}{2}}^{frac{pi}{2}} r^4cosphi mathop{dphi}mathop{dr}$$
$$= 12pi int_{0}^{a} r^4 mathop{dr}$$
$$= frac{12}{5}pi a^5$$
edited Mar 19 at 22:22
zwim
12.7k832
answered Mar 19 at 20:42
KY TangKY Tang
50436
edited Mar 19 at 22:22
zwim
12.7k832
edited Mar 19 at 22:22
zwim
12.7k832
edited Mar 19 at 22:22
zwim
12.7k832
12.7k832
answered Mar 19 at 20:42
KY TangKY Tang
50436
answered Mar 19 at 20:42
KY TangKY Tang
50436
answered Mar 19 at 20:42
KY TangKY Tang
50436
50436
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
Note you can useiint_S$iint_S$ andiiint_V$iiint_V$ for multidimensional integrals, it renders better than succession ofintint$intint$
$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
add a comment |
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
Note you can useiint_S$iint_S$ andiiint_V$iiint_V$ for multidimensional integrals, it renders better than succession ofintint$intint$
$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
You are right. I am making an edit. Thanks.
$endgroup$
– KY Tang
Mar 19 at 21:35
$begingroup$
Note you can use
iint_S $iint_S$ and iiint_V $iiint_V$ for multidimensional integrals, it renders better than succession of intint $intint$$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
Note you can use
iint_S $iint_S$ and iiint_V $iiint_V$ for multidimensional integrals, it renders better than succession of intint $intint$$endgroup$
– zwim
Mar 19 at 22:17
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
This is good to know. Thanks.
$endgroup$
– KY Tang
Mar 19 at 22:19
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
$begingroup$
+1 For Ostrogradsky :).
$endgroup$
– MachineLearner
Mar 19 at 23:52
add a comment |
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