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Math Help - Volume under parabolic surface.

  1. #1
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    Volume under parabolic surface.

    I need a check on my solutions - I don't have answers in my study guide.

    Also would appreciate commens on my 'train of thought' ie if there is a shorter way to arrive at my findings.

    2. Find the volume under the parabolic surface z=x^2+y^2 for x, y such that x^2+y^2\leq{a^2}.

    Answer.

    Define the region of integration:

    x^2+y^2\leq{a^2}from there  \frac{x^2}{a^2}+\frac{y^2}{a^2}\leq1. Therefore, the area under the function z is the ellipse with identical foci (a=b) ie cirlce.

    Since there is no one function that describes the circle, I can either split it into two parts and evaluate two integrals or - which I'd rather prefer - observe that the two areas of integration are symmetrical and are described by the functions symmetrical with relation to both X or Y axis, and only evaluate one integral, then double its value to get to the required volume.

    Is my short-cut strategy valid?...

    To evaluate one half part (one double integral), I will use variable substitution x=rcos\Theta y=rsin\Theta

    \int_0^{\pi}d\Theta\int_0^ar^2\frac{drd\Theta}{r}=  \int_0^{\pi}\frac{a^2}{2}d\Theta=\frac{\pi*a^2}{2}

    I will need to double that to find the total volume which is \pi*a^2. Looking back at the solution, however, something does not seem right as this looks too much like an area, not a volume...
    Last edited by Volga; January 5th 2011 at 12:31 AM. Reason: Moved from another thread.
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  2. #2
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    I would not call that a "variable substitution", I would say you are puttting the problem in cylindrical coordinates. That is, x and y are in polar coordinates, x= r cos(\theta), and y= r sin(\theta). Then z= r^2 is the integrand. But where did you get that "r" in the denominator? The differential of area in polar coordinates is r drd\theta, not \frac{drd\theta}{r}.

    Since the boundary of the region in the xy-plane is the circle x^2+ y^2= r^2= a^2, r ranges from 0 to 2\pi and r from 0 to a.

    The volume is given by \int_{\theta= 0}^{2\pi}\int_{r= 0}^a r^2 (r drd\theta)= \int_{\theta= 0}^{2\pi}\int_{r= 0}^a r^3 drd\theta
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  3. #3
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    Thanks! You are right, I made a silly mistake about putting r in the denominator not nominator of the integrand.

    I think I got it about boundaries of \Theta, and my concerns about 'cirlce is not a proper function' were not justified - the function in the integrand is r^3, and the cirlce is simply the area under the integrand. I probably mixed this 'cicle' thing up with an example from a different chapter ))) amazing how making mistakes make things clearer
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  4. #4
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    bounds for polar coordinates when x is not bounded

    One follow up question on using the polar coordinates:

    - if, say, my original bounds of integration were -\infty<x<+\infty, then when I use x=rcos\Theta, what are my equivalent bounds for r and Theta?

    I am thinking 0<\Theta<2\pi and 0<r<+\infty.
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  5. #5
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    That would depend upon what the bounds on y were also. Certainly, if the bounds on both x and y were -\infty < x< \infty and -\infty< y< \infty, then that would be the entire xy-plane and, in polar coordinates, 0\le \theta\le 2\pi, 0\le r< \infty which is NOT quite what you give- you have strict inequality and would not include the non-negative x-axis

    But if the bounds on x were -\infty< x< \infty and 0\le y< \infty, then you would have the upper half plane and the bounds, in polar coordinates, would be 0\le \theta\le \pi, 0\le r< \infty.
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  6. #6
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    Yes, I was thinking infinity for y also, and you've answered my question. And thanks for the correction on equality signs.
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