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Thread: 3-Dimensional Surfaces, Integrals, etc.

  1. #1
    Super Member Aryth's Avatar
    Feb 2007

    3-Dimensional Surfaces, Integrals, etc.

    Consider the inverted paraboloid defined by the equation $\displaystyle z = 10 - 2(x^2 + y^2)$. Under this area but above the x-y plane is a cylinder whose axis lies along the z-axis and whose radius is a = 2.00cm. We wish to consider the part of this cylinder that extends from the x-y plane up to the height where it intersects the paraboloid.

    a) How tall is the cylinder under the paraboloid?

    b) Consider the cylinder plus its parabolic cap (the part of the paraboloid cut out by the cylinder). What is the surface area of this curved cap?

    c) What is the entire volume contained within the cylinder, from above the x-y plane to just below the cap?

    d) The y-z plane cuts through the cylinder and its parabolic cap. What is the length of the parabolic arc formed by the cross section (that is cut out by the y-z plane) of the parabolic cap above the cylinder?

    Just some starting tips or hints would be great...
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  2. #2
    Senior Member
    Dec 2008
    a) We know that $\displaystyle z=10-2(x^2+y^2)$ may be rewritten as $\displaystyle z=10-2r^2$. When $\displaystyle r=2$, $\displaystyle z=10-2\cdot4=2$ everywhere on the circle, so the cylinder has height $\displaystyle h=2$.

    b) The surface area of the parabolic cap can be found with the formula

    $\displaystyle \int\int \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}\,dA,$

    which is derived from a cross product formula.

    c) For the volume, we may convert to polar coordinates:

    $\displaystyle V=\int_0^{2\pi}\int_0^2 (10-2r^2)\,r\,dr\,d\theta.$

    d) The length of the arc is equal to the length of $\displaystyle y=10-2x^2$ in the $\displaystyle xy$-plane from $\displaystyle x=-2$ to $\displaystyle x=2$. We may find this using the formula for arc length

    $\displaystyle s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,d x.$
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