Mixing
transforming triple integral from Cartesian to polar and solving it
Clash Royale CLAN TAG#URR8PPP
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The integral I'm trying to evaluate is the following $$iiintfrac1sqrt1-x^2-y^2-z^2dxdydz$$ taking throughout the volume of the spare $x^2-y^2-z^2=1$ lying in the first octant
Answer is $fracpi^28$
I have tried it for past a week trying to get the answer . It requires transforming it back to polar coordinate which is what I'm not able to do convincingly. I need a detailed solution. I have tried and have gotten questions similar but this is very convincing for me. I will appreciate if someone can help me out in time, it's likely it comes out in my exam next week. Thanks.
calculus
edited Jul 20 at 21:44
amWhy
189k25219431
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
add a comment |Â
up vote
-1
down vote
favorite
The integral I'm trying to evaluate is the following $$iiintfrac1sqrt1-x^2-y^2-z^2dxdydz$$ taking throughout the volume of the spare $x^2-y^2-z^2=1$ lying in the first octant
Answer is $fracpi^28$
I have tried it for past a week trying to get the answer . It requires transforming it back to polar coordinate which is what I'm not able to do convincingly. I need a detailed solution. I have tried and have gotten questions similar but this is very convincing for me. I will appreciate if someone can help me out in time, it's likely it comes out in my exam next week. Thanks.
calculus
edited Jul 20 at 21:44
amWhy
189k25219431
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Jul 20 at 21:30
Being new on MSE I want to tell you that if you find that one of the answers you get for the post solves your problem, you should mark it as accepted so that the MSE user gets the points and so that every other member on the site knows that this post is closed! It's nice, but not compulsory, even marking answers as useful if you find them so
– Davide Morgante
Jul 20 at 21:58
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
The integral I'm trying to evaluate is the following $$iiintfrac1sqrt1-x^2-y^2-z^2dxdydz$$ taking throughout the volume of the spare $x^2-y^2-z^2=1$ lying in the first octant
Answer is $fracpi^28$
I have tried it for past a week trying to get the answer . It requires transforming it back to polar coordinate which is what I'm not able to do convincingly. I need a detailed solution. I have tried and have gotten questions similar but this is very convincing for me. I will appreciate if someone can help me out in time, it's likely it comes out in my exam next week. Thanks.
calculus
edited Jul 20 at 21:44
amWhy
189k25219431
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
The integral I'm trying to evaluate is the following $$iiintfrac1sqrt1-x^2-y^2-z^2dxdydz$$ taking throughout the volume of the spare $x^2-y^2-z^2=1$ lying in the first octant
Answer is $fracpi^28$
I have tried it for past a week trying to get the answer . It requires transforming it back to polar coordinate which is what I'm not able to do convincingly. I need a detailed solution. I have tried and have gotten questions similar but this is very convincing for me. I will appreciate if someone can help me out in time, it's likely it comes out in my exam next week. Thanks.
calculus
edited Jul 20 at 21:44
amWhy
189k25219431
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
edited Jul 20 at 21:44
amWhy
189k25219431
edited Jul 20 at 21:44
amWhy
189k25219431
edited Jul 20 at 21:44
amWhy
189k25219431
189k25219431
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
asked Jul 20 at 21:25
Olaniyi Micheal mc olas
31
31
Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Jul 20 at 21:30
Being new on MSE I want to tell you that if you find that one of the answers you get for the post solves your problem, you should mark it as accepted so that the MSE user gets the points and so that every other member on the site knows that this post is closed! It's nice, but not compulsory, even marking answers as useful if you find them so
– Davide Morgante
Jul 20 at 21:58
add a comment |Â
Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Jul 20 at 21:30
Being new on MSE I want to tell you that if you find that one of the answers you get for the post solves your problem, you should mark it as accepted so that the MSE user gets the points and so that every other member on the site knows that this post is closed! It's nice, but not compulsory, even marking answers as useful if you find them so
– Davide Morgante
Jul 20 at 21:58
Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Jul 20 at 21:30
Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Jul 20 at 21:30
Being new on MSE I want to tell you that if you find that one of the answers you get for the post solves your problem, you should mark it as accepted so that the MSE user gets the points and so that every other member on the site knows that this post is closed! It's nice, but not compulsory, even marking answers as useful if you find them so
– Davide Morgante
Jul 20 at 21:58
Being new on MSE I want to tell you that if you find that one of the answers you get for the post solves your problem, you should mark it as accepted so that the MSE user gets the points and so that every other member on the site knows that this post is closed! It's nice, but not compulsory, even marking answers as useful if you find them so
– Davide Morgante
Jul 20 at 21:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
First thing first recall that $$r^2 = x^2+y^2+z^2$$ and that the Jacobian for the spherical coordinate is $$det(J) = r^2sin(theta)drdtheta dphi$$ and that $thetain[0,pi]$, $rin[0,1]$ and $phiin[0,pi]$ because we're integrating over a sphere of radius 1.
Your integral then becomes $$iiintfrac1sqrt1-(x^2+y^2+z^2)dxdydz = colorredfrac18int_0^2piint_0^piint_0^1fracr^2sin(theta)dr dtheta d phisqrt1-r^2$$
The red factor is there to take only one octant of the space (symmetry permits us to simplify our calculation like that, but we could do it by setting appropriate limits for integration).
Evaluating the integrals on the angels we get $$frac18underbrace2pi_phitext integral underbrace2_thetatext integralint_0^1fracr^2drsqrt1-r^2$$ Making a substitution $r=sin(t)$ the limits of integration becomes $$r=0 Rightarrow t = 0;;;;;r=1Rightarrow t=fracpi2$$ and $$dt = frac1sqrt1-r^2dr$$ the integral simplifies to $$fracpi2int_0^piover 2sin^2(t)dt = fracpi^28$$
answered Jul 20 at 21:49
Davide Morgante
1,812220
add a comment |Â
up vote
0
down vote
You want $frac18int_0^2pidphiint_0^pi dthetasinthetaint_0^1fracr^2 drsqrt1-r^2=fracpi2int_0^1fracr^2 drsqrt1-r^2$. The substitution $r=sin t$ lets us rewrite this as $fracpi2int_0^pi/2sin^2 t dt=fracpi2fracpi2frac12=fracpi^28$, where we have used the fact that $sin^2 t$ averages to $frac12$.
answered Jul 20 at 21:33
J.G.
13.2k11424
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First thing first recall that $$r^2 = x^2+y^2+z^2$$ and that the Jacobian for the spherical coordinate is $$det(J) = r^2sin(theta)drdtheta dphi$$ and that $thetain[0,pi]$, $rin[0,1]$ and $phiin[0,pi]$ because we're integrating over a sphere of radius 1.
Your integral then becomes $$iiintfrac1sqrt1-(x^2+y^2+z^2)dxdydz = colorredfrac18int_0^2piint_0^piint_0^1fracr^2sin(theta)dr dtheta d phisqrt1-r^2$$
The red factor is there to take only one octant of the space (symmetry permits us to simplify our calculation like that, but we could do it by setting appropriate limits for integration).
Evaluating the integrals on the angels we get $$frac18underbrace2pi_phitext integral underbrace2_thetatext integralint_0^1fracr^2drsqrt1-r^2$$ Making a substitution $r=sin(t)$ the limits of integration becomes $$r=0 Rightarrow t = 0;;;;;r=1Rightarrow t=fracpi2$$ and $$dt = frac1sqrt1-r^2dr$$ the integral simplifies to $$fracpi2int_0^piover 2sin^2(t)dt = fracpi^28$$
answered Jul 20 at 21:49
Davide Morgante
1,812220
add a comment |Â
up vote
2
down vote
accepted
First thing first recall that $$r^2 = x^2+y^2+z^2$$ and that the Jacobian for the spherical coordinate is $$det(J) = r^2sin(theta)drdtheta dphi$$ and that $thetain[0,pi]$, $rin[0,1]$ and $phiin[0,pi]$ because we're integrating over a sphere of radius 1.
Your integral then becomes $$iiintfrac1sqrt1-(x^2+y^2+z^2)dxdydz = colorredfrac18int_0^2piint_0^piint_0^1fracr^2sin(theta)dr dtheta d phisqrt1-r^2$$
The red factor is there to take only one octant of the space (symmetry permits us to simplify our calculation like that, but we could do it by setting appropriate limits for integration).
Evaluating the integrals on the angels we get $$frac18underbrace2pi_phitext integral underbrace2_thetatext integralint_0^1fracr^2drsqrt1-r^2$$ Making a substitution $r=sin(t)$ the limits of integration becomes $$r=0 Rightarrow t = 0;;;;;r=1Rightarrow t=fracpi2$$ and $$dt = frac1sqrt1-r^2dr$$ the integral simplifies to $$fracpi2int_0^piover 2sin^2(t)dt = fracpi^28$$
answered Jul 20 at 21:49
Davide Morgante
1,812220
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First thing first recall that $$r^2 = x^2+y^2+z^2$$ and that the Jacobian for the spherical coordinate is $$det(J) = r^2sin(theta)drdtheta dphi$$ and that $thetain[0,pi]$, $rin[0,1]$ and $phiin[0,pi]$ because we're integrating over a sphere of radius 1.
Your integral then becomes $$iiintfrac1sqrt1-(x^2+y^2+z^2)dxdydz = colorredfrac18int_0^2piint_0^piint_0^1fracr^2sin(theta)dr dtheta d phisqrt1-r^2$$
The red factor is there to take only one octant of the space (symmetry permits us to simplify our calculation like that, but we could do it by setting appropriate limits for integration).
Evaluating the integrals on the angels we get $$frac18underbrace2pi_phitext integral underbrace2_thetatext integralint_0^1fracr^2drsqrt1-r^2$$ Making a substitution $r=sin(t)$ the limits of integration becomes $$r=0 Rightarrow t = 0;;;;;r=1Rightarrow t=fracpi2$$ and $$dt = frac1sqrt1-r^2dr$$ the integral simplifies to $$fracpi2int_0^piover 2sin^2(t)dt = fracpi^28$$
answered Jul 20 at 21:49
Davide Morgante
1,812220
First thing first recall that $$r^2 = x^2+y^2+z^2$$ and that the Jacobian for the spherical coordinate is $$det(J) = r^2sin(theta)drdtheta dphi$$ and that $thetain[0,pi]$, $rin[0,1]$ and $phiin[0,pi]$ because we're integrating over a sphere of radius 1.
Your integral then becomes $$iiintfrac1sqrt1-(x^2+y^2+z^2)dxdydz = colorredfrac18int_0^2piint_0^piint_0^1fracr^2sin(theta)dr dtheta d phisqrt1-r^2$$
The red factor is there to take only one octant of the space (symmetry permits us to simplify our calculation like that, but we could do it by setting appropriate limits for integration).
Evaluating the integrals on the angels we get $$frac18underbrace2pi_phitext integral underbrace2_thetatext integralint_0^1fracr^2drsqrt1-r^2$$ Making a substitution $r=sin(t)$ the limits of integration becomes $$r=0 Rightarrow t = 0;;;;;r=1Rightarrow t=fracpi2$$ and $$dt = frac1sqrt1-r^2dr$$ the integral simplifies to $$fracpi2int_0^piover 2sin^2(t)dt = fracpi^28$$
answered Jul 20 at 21:49
Davide Morgante
1,812220
edited Jul 20 at 22:22
answered Jul 20 at 21:49
Davide Morgante
1,812220
answered Jul 20 at 21:49
Davide Morgante
1,812220
answered Jul 20 at 21:49
Davide Morgante
1,812220
1,812220
add a comment |Â
add a comment |Â
up vote
0
down vote
You want $frac18int_0^2pidphiint_0^pi dthetasinthetaint_0^1fracr^2 drsqrt1-r^2=fracpi2int_0^1fracr^2 drsqrt1-r^2$. The substitution $r=sin t$ lets us rewrite this as $fracpi2int_0^pi/2sin^2 t dt=fracpi2fracpi2frac12=fracpi^28$, where we have used the fact that $sin^2 t$ averages to $frac12$.
answered Jul 20 at 21:33
J.G.
13.2k11424
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
add a comment |Â
up vote
0
down vote
You want $frac18int_0^2pidphiint_0^pi dthetasinthetaint_0^1fracr^2 drsqrt1-r^2=fracpi2int_0^1fracr^2 drsqrt1-r^2$. The substitution $r=sin t$ lets us rewrite this as $fracpi2int_0^pi/2sin^2 t dt=fracpi2fracpi2frac12=fracpi^28$, where we have used the fact that $sin^2 t$ averages to $frac12$.
answered Jul 20 at 21:33
J.G.
13.2k11424
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You want $frac18int_0^2pidphiint_0^pi dthetasinthetaint_0^1fracr^2 drsqrt1-r^2=fracpi2int_0^1fracr^2 drsqrt1-r^2$. The substitution $r=sin t$ lets us rewrite this as $fracpi2int_0^pi/2sin^2 t dt=fracpi2fracpi2frac12=fracpi^28$, where we have used the fact that $sin^2 t$ averages to $frac12$.
answered Jul 20 at 21:33
J.G.
13.2k11424
You want $frac18int_0^2pidphiint_0^pi dthetasinthetaint_0^1fracr^2 drsqrt1-r^2=fracpi2int_0^1fracr^2 drsqrt1-r^2$. The substitution $r=sin t$ lets us rewrite this as $fracpi2int_0^pi/2sin^2 t dt=fracpi2fracpi2frac12=fracpi^28$, where we have used the fact that $sin^2 t$ averages to $frac12$.
answered Jul 20 at 21:33
J.G.
13.2k11424
answered Jul 20 at 21:33
J.G.
13.2k11424
answered Jul 20 at 21:33
J.G.
13.2k11424
answered Jul 20 at 21:33
J.G.
13.2k11424
13.2k11424
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
add a comment |Â
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
Thanks for the two correct answers, but I wanted to understand also why the range of Õ changes to [2pie,0] likewise the 1/8 behind, how did it get there?
– Olaniyi Micheal mc olas
Jul 23 at 7:56
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
@OlaniyiMichealmcolas As Davide Morgante noted, the $1/8$ factor is due to restriction to the positive octant, and each octant having the same contribution to an all-octants integral due to symmetry. If we had wanted all octants, the ranges of $theta,,phi$ would respectively be $[0,,pi],,0,,2pi]$; this is standard in the spherical polar coordinate system.
– J.G.
Jul 23 at 8:02
add a comment |Â
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Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Jul 20 at 21:30
Being new on MSE I want to tell you that if you find that one of the answers you get for the post solves your problem, you should mark it as accepted so that the MSE user gets the points and so that every other member on the site knows that this post is closed! It's nice, but not compulsory, even marking answers as useful if you find them so
– Davide Morgante
Jul 20 at 21:58