1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Volume of the solid enclosed by the paraboloids

  1. Oct 28, 2005 #1
    Find the volume of the solid enclosed by the paraboloids z= 16(x^2 +y^2) and z=32-16(x^2+y^2)

    i'm not sure how i would find the x bounds for this triple integral. here's my work:

    16x^2+16y^2 = 32-16x^2+16y^2 => simplifies to y = +- sqrt(1-x^2) (the y-bounds)

    z bounds is already given which is from z= 16x^2+16y^2 to 32-16x^2+16y^2

    but how would i find the x bounds (mathematically if possible)? i'm not sure
  2. jcsd
  3. Oct 28, 2005 #2
    I would try a different coordinate system?
  4. Oct 28, 2005 #3


    User Avatar
    Science Advisor

    Adding the two equations gives 2z= 32 or z= 16. When z= 16 both equations reduce to x^2+ y^2= 1 so you are correct. x ranges from -1 to 1, for each x, y ranges from [itex]-\sqrt{1-x^2}[/itex] to [itex]\sqrt{1-x^2}[/itex], for each x,y, z ranges from 16(x2+ y[/sup]2[/sup]) to
    32- 16(x2+ y2.
    The volume is
    [tex]\int_{x= -1}^1 \int{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{z=16(x^2+ y^2)}^{32-16(x^2+ y^2} dzdydx[/tex]
    Yes, you would find it simpler to take advantage of the circular symmetry by changing to cylindrical coordinates: r, [itex]\theta[/itex], z.
    The largest extent parallel to the xy-plane is at the intersection of the two parabolas, a circle with radius 1 so r ranges from 0 to 1, [itex]\theta[/itex] varies from 0 to [itex]2\pi[/tex] and z varies from 16r2 to 32- 16r2.
    The volume is
    [tex]\int_{r=0}^1 \int_{\theta=0}^{2\pi} \int_{z= 15r^2}^1 r dzd\thetadr[/tex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook