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Homework Help: Finding volume between two surfaces.

  1. Apr 21, 2012 #1
    I am having trouble finding the volume between the two surfaces. Well I did the problem and got an answer but im not sure if im approaching this type of problem correctly. So can you take a look at how i approached the problem and tell me if i am on the right path or if it is incorrect. Thank you in advance.

    Find the volume between the two surfaces
    z = x^2 + 3y^2
    z = 8 - x^2 - y^2

    First I set both equations equal to each other to get y =±sqrt(2 - x^2).
    Finding the volume will require a double integral so this can be my y limits.
    X cannot exceed sqrt(2) so i assume my x limits are ±sqrt(2)

    The sum of the two functions is 2y^2 + 8. I believe this is what i should integrate. If not please tell me what i should integrate and why.

    So i integrate 2y^2 + 8 with respect to y and get (2/3)y^3 + 8y. And inputting the y limits yields (4/3)*(2 - x^2)^(3/2) + 16*(2 - x^2)^(1/2).

    Now i integrate that with respect to x using the computer and get
    18*arcsin(sqrt(2)x/2) - (1/3)x*(x^2 - 5)*sqrt(2 - x^2) + 8x*sqrt(2 - x^2)

    Putting in limits of ±sqrt(2) gets a final answer of 18*pi.
  2. jcsd
  3. Apr 21, 2012 #2


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    Well, you shouldn't get that! Did you forget the "3" multiplying [itex]y^2[/itex] in the first equation? [itex]z= x^2+ 3y^2= 8- x^2- y^2[/itex] gives [itex]2x^2+ 4y^2= 8[/itex]
    or [itex]x^2+ 2y^2= 4[/itex]. That's an ellipse. Solving for y, [itex]y= \pm(1/2)\sqrt{4- x^2}[/itex].

    If the "ceiling" is given by f(x,y) and the "floor" by g(x,y) the "height" is f- g, not f+ g. Subtract, don't add! Here, the "ceiling" is [itex]f(x,y)= 8- x^2- y^2[/itex] and the "floor" is [itex]g(x,y)= x^2+ 3y^2[/itex]. The distance between them, the height, is [itex](8- x^2- y^2)- (x^2+ 3y^2)= 8- 2x^2- 4y^2[/itex].

    As above, x goes from -2 to 2 and, for each x, y goes from [itex]-(1/2)\sqrt{4- x^2}[/itex] to [itex](1/2)\sqrt{4- x^2}[/itex].
    [tex]\int_{x=-2}^2\int_{-(1/2)\sqrt{4- x^2}}^{(1/2)\sqrt{4- x^2}}8- 2x^2- 4y^2 dydx[/tex]

    Another way to do this is to write the ellipse, at which the two paraboloids intersect, as
    [tex]\frac{x^2}{4}+ \frac{y^2}{2}= 1[/tex]

    Let [itex]x= 2rcos(\theta)[/itex], [itex]y= \sqrt{2}rsin(\theta)[/itex] so that the Jacobian is
    [tex]\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta}\end{array}\right|= \left|\begin{array}{cc}2cos(t) & -2rsin(\theta) \\ \sqrt{2}sin(\theta) & \sqrt{2}rcos(\theta) \end{array}\right|= 2\sqrt{2}r[/tex]
    And the integrand is [itex]8- 2x^2- 4y^2= 8- 4r^2cos^2(\theta)- 8r^2sin^2(\theta)[/itex]. You can use the fact that [itex]sin^2(\theta)+ cos^2(\theta)[/itex] to write this as [itex]8- 4r^2(cos^2(\theta)+ sin^2(\theta))- 4r^2sin^2(\theta)= 4- 4r^2sin^2(\theta)[/itex]
    so the integral would be
    [tex]\int_{r=0}^1\int_{\theta= 0}^{2\pi} (4- 4r^2sin^2(\theta))(2\sqrt{2} r dr d\theta)[/tex]
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