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    Find the volume within the cylinder r = 4 cos (theta) bounded below the sphere r^2+ z^2 = 16 and above by the plane z = 0

    \int_{}^{} \int_{}^{} \int_{0}^{\sqrt{16-r^2}} r  dzdrd\theta

    I have the above so far,,, i'm trying to project the shadow onto the x-y plane but don't know how to do it any help?

    From here I can do the rest but dunno how to get to the shadow

    Thanks
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    (Original post by CHEN20041)
    Find the volume within the cylinder r = 4 cos (theta) bounded below the sphere r^2+ z^2 = 16 and above by the plane z = 0

    \int_{}^{} \int_{}^{} \int_{0}^{\sqrt{16-r^2}} r  dzdrd\theta

    I have the above so far,,, i'm trying to project the shadow onto the x-y plane but don't know how to do it any help?

    From here I can do the rest but dunno how to get to the shadow

    Thanks
    Why? Just fill in the missing limits and calculate it appropriately to answer the question.
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    (Original post by RDKGames)
    Why? Just fill in the missing limits and calculate it appropriately to answer the question.
    Yeah I guess you're right

    \int_{0}^{\pi} \int_{0}^{4cos\theta} \int_{0}^{\sqrt{16-r^2}} r dzdrd\theta

    this is the answer in the mark scheme, why is the limits for theta taken as pi.0 and not 2pi,0 I mean it's a sphere so full rotation around the axis no?
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    (Original post by CHEN20041)
    Yeah I guess you're right

    \int_{0}^{\pi} \int_{0}^{4cos\theta} \int_{0}^{\sqrt{16-r^2}} r dzdrd\theta

    this is the answer in the mark scheme, why is the limits for theta taken as pi.0 and not 2pi,0 I mean it's a sphere so full rotation around the axis no?
    But you don't want a full sphere though, you want half of it due to the z>0 restriction.
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    (Original post by RDKGames)
    But you don't want a full sphere though, you want half of it due to the z>0 restriction.
    Yeah that's what I was thinking but i'm confused since I was doing the following question which also uses half a sphere but the limits for theta are 2pi,0

    \int_{}^{} \int_{}^{} \int_{}^{} 16zdv where the region R is upper half of unit sphere.

    The bounds for theta for this was taken to be 2pi,0 but that was using spherical coordinates dunno if that makes a difference.
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    (Original post by CHEN20041)
    Yeah that's what I was thinking but i'm confused since I was doing the following question which also uses half a sphere but the limits for theta are 2pi,0

    \int_{}^{} \int_{}^{} \int_{}^{} 16zdv where the region R is upper half of unit sphere.

    The bounds for theta for this was taken to be 2pi,0 but that was using spherical coordinates dunno if that makes a difference.
    The other angle must've been halved then to \frac{\pi}{2}, because there are two angles in spherical coords.
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    (Original post by RDKGames)
    The other angle must've been halved then to \frac{\pi}{2}, because there are two angles in spherical coords.
    Yeah it was

    pi/2 and 0 ... but that is measuring going from z-axis right?

    Still don't get why the theta angle would be 2pi,0 since it's the same angle as it is in cylindrical coords?
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    (Original post by CHEN20041)
    Yeah it was

    pi/2 and 0 ... but that is measuring going from z-axis right?

    Still don't get why the theta angle would be 2pi,0 since it's the same angle as it is in cylindrical coords?
    It depends how you define them, since they can be interchanged.

    Dunno how to explain it without some visualisation of 3D space and the sphere involved, but if you go here: https://en.wikipedia.org/wiki/Spheri...rdinate_system

    ...and look at the second diagram on the right (the one which is used more often in maths, as the description says). Then notice that the angle with domain [0,180] is measured AROUND the z-axis from the x-axis. Not FROM the z-axis.


    As far as your Q is concerned, \theta is the same as in the above system, measured around the z-axis, however in spherical it doesn't require to go a full 360 cycle, only the 180. Whereas in cylindrical, you need to go a full 360 to define a full cylinder (or sphere, as above in your Q). So hence, you halve that here.
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    (Original post by RDKGames)
    It depends how you define them, since they can be interchanged.

    Dunno how to explain it without some visualisation of 3D space and the sphere involved, but if you go here: https://en.wikipedia.org/wiki/Spheri...rdinate_system

    ...and look at the second diagram on the right (the one which is used more often in maths, as the description says). Then notice that the angle with domain [0,180] is measured AROUND the z-axis from the x-axis. Not FROM the z-axis.


    As far as your Q is concerned, \theta is the same as in the above system, measured around the z-axis, however in spherical it doesn't require to go a full 360 cycle, only the 180. Whereas in cylindrical, you need to go a full 360 to define a full cylinder (or sphere, as above in your Q). So hence, you halve that here.
    Ohhh I see that make it clearer, thanks for that
 
 
 
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