Area of the part of the sphere $x^2+y^2+z^2=36$ limited by the cylinder $x^2+y^2=6y$
I have
$$sqrt{1+bigg(frac{partial z}{partial x}bigg)^2+bigg(frac{partial z}{partial y}bigg)^2}=frac{6}{sqrt{36-x^2-y^2}}$$
where $z=sqrt{36-x^2-y^2}$
In the polar form I have to solve
$$frac{sigma}{2}= int_0^pi int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=int_0^pibigg[-6sqrt{36-rho^2}bigg]_0^{6sin theta},dtheta=bigg[36theta -36sin thetabigg]_0^pi=^?36pi$$
If I solve this way
$$frac{sigma}{4}= int_0^{frac{pi}{2}} int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=18pi-36$$
It gives me the real answer $sigma=72pi-144$
Why this happens?
integration multivariable-calculus
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
add a comment |
I have
$$sqrt{1+bigg(frac{partial z}{partial x}bigg)^2+bigg(frac{partial z}{partial y}bigg)^2}=frac{6}{sqrt{36-x^2-y^2}}$$
where $z=sqrt{36-x^2-y^2}$
In the polar form I have to solve
$$frac{sigma}{2}= int_0^pi int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=int_0^pibigg[-6sqrt{36-rho^2}bigg]_0^{6sin theta},dtheta=bigg[36theta -36sin thetabigg]_0^pi=^?36pi$$
If I solve this way
$$frac{sigma}{4}= int_0^{frac{pi}{2}} int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=18pi-36$$
It gives me the real answer $sigma=72pi-144$
Why this happens?
integration multivariable-calculus
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
I put $=^?$ because WolframAlpha gives the real answer to the same double integral
– Mauricio J. S.
Nov 28 '18 at 15:33
I'm confused...is the cylinder $x^2 + y^2 = 6$ or $x^2 + y^2 = 6y$. If it's the latter, then I don't see how your inner integral has the correct bounds. (It might well have the correct bounds...I just don't see it!)
– John Hughes
Nov 28 '18 at 15:34
I have a mistake, I'll fix immediately
– Mauricio J. S.
Nov 28 '18 at 15:38
add a comment |
I have
$$sqrt{1+bigg(frac{partial z}{partial x}bigg)^2+bigg(frac{partial z}{partial y}bigg)^2}=frac{6}{sqrt{36-x^2-y^2}}$$
where $z=sqrt{36-x^2-y^2}$
In the polar form I have to solve
$$frac{sigma}{2}= int_0^pi int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=int_0^pibigg[-6sqrt{36-rho^2}bigg]_0^{6sin theta},dtheta=bigg[36theta -36sin thetabigg]_0^pi=^?36pi$$
If I solve this way
$$frac{sigma}{4}= int_0^{frac{pi}{2}} int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=18pi-36$$
It gives me the real answer $sigma=72pi-144$
Why this happens?
integration multivariable-calculus
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
I have
$$sqrt{1+bigg(frac{partial z}{partial x}bigg)^2+bigg(frac{partial z}{partial y}bigg)^2}=frac{6}{sqrt{36-x^2-y^2}}$$
where $z=sqrt{36-x^2-y^2}$
In the polar form I have to solve
$$frac{sigma}{2}= int_0^pi int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=int_0^pibigg[-6sqrt{36-rho^2}bigg]_0^{6sin theta},dtheta=bigg[36theta -36sin thetabigg]_0^pi=^?36pi$$
If I solve this way
$$frac{sigma}{4}= int_0^{frac{pi}{2}} int_0^{6sin theta} frac{6}{sqrt{36-rho^2}}rho,drho,dtheta=18pi-36$$
It gives me the real answer $sigma=72pi-144$
Why this happens?
integration multivariable-calculus
integration multivariable-calculus
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
edited Nov 28 '18 at 15:40
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
asked Nov 28 '18 at 15:31
Mauricio J. S.
204
204
I put $=^?$ because WolframAlpha gives the real answer to the same double integral
– Mauricio J. S.
Nov 28 '18 at 15:33
I'm confused...is the cylinder $x^2 + y^2 = 6$ or $x^2 + y^2 = 6y$. If it's the latter, then I don't see how your inner integral has the correct bounds. (It might well have the correct bounds...I just don't see it!)
– John Hughes
Nov 28 '18 at 15:34
I have a mistake, I'll fix immediately
– Mauricio J. S.
Nov 28 '18 at 15:38
add a comment |
I put $=^?$ because WolframAlpha gives the real answer to the same double integral
– Mauricio J. S.
Nov 28 '18 at 15:33
I'm confused...is the cylinder $x^2 + y^2 = 6$ or $x^2 + y^2 = 6y$. If it's the latter, then I don't see how your inner integral has the correct bounds. (It might well have the correct bounds...I just don't see it!)
– John Hughes
Nov 28 '18 at 15:34
I have a mistake, I'll fix immediately
– Mauricio J. S.
Nov 28 '18 at 15:38
I put $=^?$ because WolframAlpha gives the real answer to the same double integral
– Mauricio J. S.
Nov 28 '18 at 15:33
I put $=^?$ because WolframAlpha gives the real answer to the same double integral
– Mauricio J. S.
Nov 28 '18 at 15:33
I'm confused...is the cylinder $x^2 + y^2 = 6$ or $x^2 + y^2 = 6y$. If it's the latter, then I don't see how your inner integral has the correct bounds. (It might well have the correct bounds...I just don't see it!)
– John Hughes
Nov 28 '18 at 15:34
I'm confused...is the cylinder $x^2 + y^2 = 6$ or $x^2 + y^2 = 6y$. If it's the latter, then I don't see how your inner integral has the correct bounds. (It might well have the correct bounds...I just don't see it!)
– John Hughes
Nov 28 '18 at 15:34
I have a mistake, I'll fix immediately
– Mauricio J. S.
Nov 28 '18 at 15:38
I have a mistake, I'll fix immediately
– Mauricio J. S.
Nov 28 '18 at 15:38
add a comment |
1 Answer
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You simplified a square root incorrectly. When you substitute the limits $rho=0$ and $rho=6sintheta$ to the function $sqrt{36-rho^2}$ at the upper limit you get
$$
sqrt{36-36sin^2theta}=sqrt{36cos^2theta}=6|cos theta|.
$$
It seems to me that you forgot to take the absolute value. Because $costheta$ is negative in the interval $thetain(pi/2,pi]$ this makes a difference.
With the absolute value in place you are more or less forced to do the subintervals $thetain[0,pi/2]$ and $thetain[pi/2,pi]$ separately leading to your other solution.
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
1
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
add a comment |
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1 Answer
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You simplified a square root incorrectly. When you substitute the limits $rho=0$ and $rho=6sintheta$ to the function $sqrt{36-rho^2}$ at the upper limit you get
$$
sqrt{36-36sin^2theta}=sqrt{36cos^2theta}=6|cos theta|.
$$
It seems to me that you forgot to take the absolute value. Because $costheta$ is negative in the interval $thetain(pi/2,pi]$ this makes a difference.
With the absolute value in place you are more or less forced to do the subintervals $thetain[0,pi/2]$ and $thetain[pi/2,pi]$ separately leading to your other solution.
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
1
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
add a comment |
You simplified a square root incorrectly. When you substitute the limits $rho=0$ and $rho=6sintheta$ to the function $sqrt{36-rho^2}$ at the upper limit you get
$$
sqrt{36-36sin^2theta}=sqrt{36cos^2theta}=6|cos theta|.
$$
It seems to me that you forgot to take the absolute value. Because $costheta$ is negative in the interval $thetain(pi/2,pi]$ this makes a difference.
With the absolute value in place you are more or less forced to do the subintervals $thetain[0,pi/2]$ and $thetain[pi/2,pi]$ separately leading to your other solution.
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
1
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
add a comment |
You simplified a square root incorrectly. When you substitute the limits $rho=0$ and $rho=6sintheta$ to the function $sqrt{36-rho^2}$ at the upper limit you get
$$
sqrt{36-36sin^2theta}=sqrt{36cos^2theta}=6|cos theta|.
$$
It seems to me that you forgot to take the absolute value. Because $costheta$ is negative in the interval $thetain(pi/2,pi]$ this makes a difference.
With the absolute value in place you are more or less forced to do the subintervals $thetain[0,pi/2]$ and $thetain[pi/2,pi]$ separately leading to your other solution.
You simplified a square root incorrectly. When you substitute the limits $rho=0$ and $rho=6sintheta$ to the function $sqrt{36-rho^2}$ at the upper limit you get
$$
sqrt{36-36sin^2theta}=sqrt{36cos^2theta}=6|cos theta|.
$$
It seems to me that you forgot to take the absolute value. Because $costheta$ is negative in the interval $thetain(pi/2,pi]$ this makes a difference.
With the absolute value in place you are more or less forced to do the subintervals $thetain[0,pi/2]$ and $thetain[pi/2,pi]$ separately leading to your other solution.
edited Nov 28 '18 at 18:49
community wiki
2 revs
Jyrki Lahtonen
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
1
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
add a comment |
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
1
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
How I can know when I have to put the absolute value? thanks very much your answer, it is very useful.
– Mauricio J. S.
Nov 28 '18 at 16:50
1
1
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
@MauricioJ.S. The rule that is always valid reads $$sqrt{x^2}=|x|.$$ Whether you can drop the absolute value sign depends on whether $x$ can be negative. Here it can.
– Jyrki Lahtonen
Nov 28 '18 at 18:47
add a comment |
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I put $=^?$ because WolframAlpha gives the real answer to the same double integral
– Mauricio J. S.
Nov 28 '18 at 15:33
I'm confused...is the cylinder $x^2 + y^2 = 6$ or $x^2 + y^2 = 6y$. If it's the latter, then I don't see how your inner integral has the correct bounds. (It might well have the correct bounds...I just don't see it!)
– John Hughes
Nov 28 '18 at 15:34
I have a mistake, I'll fix immediately
– Mauricio J. S.
Nov 28 '18 at 15:38