Volume inside sphere bounded by plane in double integrals












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I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.










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  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 '18 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 '18 at 8:36
















0














I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.










share|cite|improve this question
























  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 '18 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 '18 at 8:36














0












0








0







I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.










share|cite|improve this question















I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.







integration multivariable-calculus polar-coordinates






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edited Nov 22 '18 at 8:41

























asked Nov 20 '18 at 15:19









mstro

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11












  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 '18 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 '18 at 8:36


















  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 '18 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 '18 at 8:36
















Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 '18 at 16:27




Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 '18 at 16:27












I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 '18 at 8:36




I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 '18 at 8:36










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The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






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    The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
    Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
    It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
    $$frac{1}{3}pi(9-c^2)c$$
    The second is the volume between the cone and z = 0 which can be obtained by
    $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
    Then the total volume is
    $$pi(9c - frac{c^3}{3})$$
    In fact it is easier to use xyz co-ordinates.
    $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






    share|cite|improve this answer


























      0














      The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
      Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
      It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
      $$frac{1}{3}pi(9-c^2)c$$
      The second is the volume between the cone and z = 0 which can be obtained by
      $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
      Then the total volume is
      $$pi(9c - frac{c^3}{3})$$
      In fact it is easier to use xyz co-ordinates.
      $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






      share|cite|improve this answer
























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        The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
        Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
        It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
        $$frac{1}{3}pi(9-c^2)c$$
        The second is the volume between the cone and z = 0 which can be obtained by
        $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
        Then the total volume is
        $$pi(9c - frac{c^3}{3})$$
        In fact it is easier to use xyz co-ordinates.
        $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






        share|cite|improve this answer












        The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
        Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
        It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
        $$frac{1}{3}pi(9-c^2)c$$
        The second is the volume between the cone and z = 0 which can be obtained by
        $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
        Then the total volume is
        $$pi(9c - frac{c^3}{3})$$
        In fact it is easier to use xyz co-ordinates.
        $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 '18 at 22:47









        KY Tang

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