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How to solve the integral of $int_-pi/2^pi/2 int_0 ^costheta sqrt12r^6 sin^2 (2theta)$

How to integrate $int_0^1 int_0^pi int_0^pi r^2 sintheta sqrt1 - r^2cos^2theta - r^2sin^2theta ,dphi, dtheta ,dr$Evaluating the surface integral side of a divergence theorem problem$int_-pi/2^pi/2 cos(a costheta) e^imtheta e^-ibsintheta mathrmdtheta $ IntegrationHow to integrate $e^-cos(theta)cos(theta + sin(theta))$Definite Integral $ 4piint_0^1cosh(t)sqrtcosh^2(t)+sinh^2(t) dt $Convert the integral from rectangular to cylindrical coordinates and solveHow to reverse the integration order of the double integral $int_theta=0^2piint_r=0^1+costhetar^2(sintheta+costheta)drdtheta$.Triple integral $int_0^2pi int_0^2cos(theta) int_0^sqrt2rcos(theta) r dzdrdtheta$ to find volume of a solidStokes theorem on surface integralHow to evaluate the integral $int_0^2pi thetaexp(xcos(theta) + ysin(theta))) dtheta$

-1

1

$begingroup$

So what I am trying to do is to find the surface area over a disc inside a sphere. The sphere is given by: $x^2 + y^2 + z^2 = 1$ and the disc: $(x-(frac12))^2 +y^2 le frac14$

I found the boundaries to be $frac-pi2 le thetale fracpi2$ and $0le r le costheta $

So now I am trying to solve $ int_-pi/2^pi/2 int_0 ^costheta dS$

Where $dS = leftlvert fracdmathbfRdr times fracdmathbfRdtheta rightrvert$

I've parameterized the sphere as
$ x=r cos(theta) , y = r sin(theta) and z=sqrt1-r^2$

Getting $ mathbfR$ = (($r cos(theta)^2)$,($r sin(theta)^2)$,($1-r^2$))

Solving $fracdmathbfRdr$ gives me $(2rcos^2 theta, 2rsin^2 theta, -2r)$ and $fracdmathbfRdtheta$ gives me $(-r^2 sin(2theta),(r^2 sin(2theta),0)$

Soving $dS$ gets me $sqrt12r^6 sin^2 (2theta)$

So now I am struggling to find a way to calculate the integral of: $int_-pi/2^pi/2 int_0 ^costheta sqrt12r^6 sin^2(2theta) dr dtheta $

integration multivariable-calculus definite-integrals

share|cite|improve this question

edited 17 hours ago

LostCoder

asked yesterday

LostCoderLostCoder

145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're missing }
  $endgroup$
  – J. W. Tanner
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Don't you think that a d$theta$d$r$ would be welcome ?
  $endgroup$
  – Jean Marie
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  and you could write pi to get $pi$
  $endgroup$
  – J. W. Tanner
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am very sorry for the inconvenience, really tired and new to Math Jax, thanks for the corrections
  $endgroup$
  – LostCoder
  yesterday
  

  
  
  
  

add a comment | 

-1

1

$begingroup$

So what I am trying to do is to find the surface area over a disc inside a sphere. The sphere is given by: $x^2 + y^2 + z^2 = 1$ and the disc: $(x-(frac12))^2 +y^2 le frac14$

I found the boundaries to be $frac-pi2 le thetale fracpi2$ and $0le r le costheta $

So now I am trying to solve $ int_-pi/2^pi/2 int_0 ^costheta dS$

Where $dS = leftlvert fracdmathbfRdr times fracdmathbfRdtheta rightrvert$

I've parameterized the sphere as
$ x=r cos(theta) , y = r sin(theta) and z=sqrt1-r^2$

Getting $ mathbfR$ = (($r cos(theta)^2)$,($r sin(theta)^2)$,($1-r^2$))

Solving $fracdmathbfRdr$ gives me $(2rcos^2 theta, 2rsin^2 theta, -2r)$ and $fracdmathbfRdtheta$ gives me $(-r^2 sin(2theta),(r^2 sin(2theta),0)$

Soving $dS$ gets me $sqrt12r^6 sin^2 (2theta)$

So now I am struggling to find a way to calculate the integral of: $int_-pi/2^pi/2 int_0 ^costheta sqrt12r^6 sin^2(2theta) dr dtheta $

integration multivariable-calculus definite-integrals

share|cite|improve this question

edited 17 hours ago

LostCoder

asked yesterday

LostCoderLostCoder

145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're missing }
  $endgroup$
  – J. W. Tanner
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Don't you think that a d$theta$d$r$ would be welcome ?
  $endgroup$
  – Jean Marie
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  and you could write pi to get $pi$
  $endgroup$
  – J. W. Tanner
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am very sorry for the inconvenience, really tired and new to Math Jax, thanks for the corrections
  $endgroup$
  – LostCoder
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

So what I am trying to do is to find the surface area over a disc inside a sphere. The sphere is given by: $x^2 + y^2 + z^2 = 1$ and the disc: $(x-(frac12))^2 +y^2 le frac14$

I found the boundaries to be $frac-pi2 le thetale fracpi2$ and $0le r le costheta $

So now I am trying to solve $ int_-pi/2^pi/2 int_0 ^costheta dS$

Where $dS = leftlvert fracdmathbfRdr times fracdmathbfRdtheta rightrvert$

I've parameterized the sphere as
$ x=r cos(theta) , y = r sin(theta) and z=sqrt1-r^2$

Getting $ mathbfR$ = (($r cos(theta)^2)$,($r sin(theta)^2)$,($1-r^2$))

Solving $fracdmathbfRdr$ gives me $(2rcos^2 theta, 2rsin^2 theta, -2r)$ and $fracdmathbfRdtheta$ gives me $(-r^2 sin(2theta),(r^2 sin(2theta),0)$

Soving $dS$ gets me $sqrt12r^6 sin^2 (2theta)$

So now I am struggling to find a way to calculate the integral of: $int_-pi/2^pi/2 int_0 ^costheta sqrt12r^6 sin^2(2theta) dr dtheta $

integration multivariable-calculus definite-integrals

share|cite|improve this question

edited 17 hours ago

LostCoder

asked yesterday

LostCoderLostCoder

145

$endgroup$

share|cite|improve this question

edited 17 hours ago

LostCoder

asked yesterday

LostCoderLostCoder

145

share|cite|improve this question

edited 17 hours ago

LostCoder

asked yesterday

LostCoderLostCoder

145

asked yesterday

LostCoderLostCoder

145

asked yesterday

LostCoderLostCoder

145

2 Answers
2

active

oldest

votes

1

$begingroup$

Step 1: the integrand is even, so I would integrate $theta$ from $0$ to $pi/2$. This way everything under the square root is a positive quantity, so the integral becomes $$2int_0^pi/2sintheta dthetaint_0^costhetadrsqrt12r^3$$
Once you integrate over $r$, you will get an expression in terms of $costheta$. Make the substitution $u=costheta$, with $du=-sintheta dtheta$. The limits of integration become $cos 0=1$ (lower limit) and $cosfrac pi2=0$ (upper limit). Note that changing the order of limits means an extra minus sign, that will cancel the one from $du$.

share|cite|improve this answer

answered yesterday

AndreiAndrei

13.1k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry if this is a stupid question, but I don't follow/understand the first step, what's happening there?
  $endgroup$
  – LostCoder
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f(x)=f(-x)$, an integral $int_-a^af(x)dx$ can be written as $int_-a^0f(x)dx+int_0^af(x)dx=-int_0^-af(x)dx+int_0^af(x)dx$. Now changing the variable in the first integral from $x$ to $-x$ you get $-int_0^-af(x)dx+int_0^af(x)dx=int_0^af(x)dx+int_0^af(x)dx=2int_0^af(x)dx$.
  $endgroup$
  – Andrei
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – LostCoder
  22 hours ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

Mathematica says

Integrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$fracsqrt35 $$

NIntegrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$0.34641 $$

Addition. The Mathematica answer to the edited by OP question is

Integrate[r^3*Sqrt[12*Sin[2*t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$frac1sqrt3 $

share|cite|improve this answer

edited 10 hours ago

answered 18 hours ago

user64494user64494

2,9651032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looking back it should be $sqrt12r^6 sin^2 (2theta)$ , however either way I am looking for a way to solve this by hand
  $endgroup$
  – LostCoder
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LostCoder: I added the answer to the question edited by you.
  $endgroup$
  – user64494
  10 hours ago
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Step 1: the integrand is even, so I would integrate $theta$ from $0$ to $pi/2$. This way everything under the square root is a positive quantity, so the integral becomes $$2int_0^pi/2sintheta dthetaint_0^costhetadrsqrt12r^3$$
Once you integrate over $r$, you will get an expression in terms of $costheta$. Make the substitution $u=costheta$, with $du=-sintheta dtheta$. The limits of integration become $cos 0=1$ (lower limit) and $cosfrac pi2=0$ (upper limit). Note that changing the order of limits means an extra minus sign, that will cancel the one from $du$.

share|cite|improve this answer

answered yesterday

AndreiAndrei

13.1k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry if this is a stupid question, but I don't follow/understand the first step, what's happening there?
  $endgroup$
  – LostCoder
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f(x)=f(-x)$, an integral $int_-a^af(x)dx$ can be written as $int_-a^0f(x)dx+int_0^af(x)dx=-int_0^-af(x)dx+int_0^af(x)dx$. Now changing the variable in the first integral from $x$ to $-x$ you get $-int_0^-af(x)dx+int_0^af(x)dx=int_0^af(x)dx+int_0^af(x)dx=2int_0^af(x)dx$.
  $endgroup$
  – Andrei
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – LostCoder
  22 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Step 1: the integrand is even, so I would integrate $theta$ from $0$ to $pi/2$. This way everything under the square root is a positive quantity, so the integral becomes $$2int_0^pi/2sintheta dthetaint_0^costhetadrsqrt12r^3$$
Once you integrate over $r$, you will get an expression in terms of $costheta$. Make the substitution $u=costheta$, with $du=-sintheta dtheta$. The limits of integration become $cos 0=1$ (lower limit) and $cosfrac pi2=0$ (upper limit). Note that changing the order of limits means an extra minus sign, that will cancel the one from $du$.

share|cite|improve this answer

answered yesterday

AndreiAndrei

13.1k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry if this is a stupid question, but I don't follow/understand the first step, what's happening there?
  $endgroup$
  – LostCoder
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f(x)=f(-x)$, an integral $int_-a^af(x)dx$ can be written as $int_-a^0f(x)dx+int_0^af(x)dx=-int_0^-af(x)dx+int_0^af(x)dx$. Now changing the variable in the first integral from $x$ to $-x$ you get $-int_0^-af(x)dx+int_0^af(x)dx=int_0^af(x)dx+int_0^af(x)dx=2int_0^af(x)dx$.
  $endgroup$
  – Andrei
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – LostCoder
  22 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Step 1: the integrand is even, so I would integrate $theta$ from $0$ to $pi/2$. This way everything under the square root is a positive quantity, so the integral becomes $$2int_0^pi/2sintheta dthetaint_0^costhetadrsqrt12r^3$$
Once you integrate over $r$, you will get an expression in terms of $costheta$. Make the substitution $u=costheta$, with $du=-sintheta dtheta$. The limits of integration become $cos 0=1$ (lower limit) and $cosfrac pi2=0$ (upper limit). Note that changing the order of limits means an extra minus sign, that will cancel the one from $du$.

share|cite|improve this answer

answered yesterday

AndreiAndrei

13.1k21230

$endgroup$

share|cite|improve this answer

answered yesterday

AndreiAndrei

13.1k21230

answered yesterday

AndreiAndrei

13.1k21230

answered yesterday

AndreiAndrei

13.1k21230

-1

$begingroup$

Mathematica says

Integrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$fracsqrt35 $$

NIntegrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$0.34641 $$

Addition. The Mathematica answer to the edited by OP question is

Integrate[r^3*Sqrt[12*Sin[2*t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$frac1sqrt3 $

share|cite|improve this answer

edited 10 hours ago

answered 18 hours ago

user64494user64494

2,9651032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looking back it should be $sqrt12r^6 sin^2 (2theta)$ , however either way I am looking for a way to solve this by hand
  $endgroup$
  – LostCoder
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LostCoder: I added the answer to the question edited by you.
  $endgroup$
  – user64494
  10 hours ago
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

Mathematica says

Integrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$fracsqrt35 $$

NIntegrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$0.34641 $$

Addition. The Mathematica answer to the edited by OP question is

Integrate[r^3*Sqrt[12*Sin[2*t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$frac1sqrt3 $

share|cite|improve this answer

edited 10 hours ago

answered 18 hours ago

user64494user64494

2,9651032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looking back it should be $sqrt12r^6 sin^2 (2theta)$ , however either way I am looking for a way to solve this by hand
  $endgroup$
  – LostCoder
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LostCoder: I added the answer to the question edited by you.
  $endgroup$
  – user64494
  10 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Mathematica says

Integrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$fracsqrt35 $$

NIntegrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$$0.34641 $$

Addition. The Mathematica answer to the edited by OP question is

Integrate[r^3*Sqrt[12*Sin[2*t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

$frac1sqrt3 $

share|cite|improve this answer

edited 10 hours ago

answered 18 hours ago

user64494user64494

2,9651032

$endgroup$

Integrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

NIntegrate[Sqrt[12*r^6*Sin[t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

Integrate[r^3*Sqrt[12*Sin[2*t]^2], t, -Pi/2, Pi/2, r, 0, Cos[t]]

share|cite|improve this answer

edited 10 hours ago

answered 18 hours ago

user64494user64494

2,9651032

answered 18 hours ago

user64494user64494

2,9651032

answered 18 hours ago

user64494user64494

2,9651032