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Thread: 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 $\displaystyle z=x^2+y^2$ for x, y such that $\displaystyle x^2+y^2\leq{a^2}$.

    Answer.

    Define the region of integration:

    $\displaystyle x^2+y^2\leq{a^2}$from there $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{a^2}\leq1$. 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 $\displaystyle x=rcos\Theta y=rsin\Theta$

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

    I will need to double that to find the total volume which is $\displaystyle \pi*a^2$. 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; Jan 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, $\displaystyle x= r cos(\theta)$, and $\displaystyle y= r sin(\theta)$. 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 $\displaystyle r drd\theta$, not $\displaystyle \frac{drd\theta}{r}$.

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

    The volume is given by $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^a r^2 (r drd\theta)= \int_{\theta= 0}^{2\pi}\int_{r= 0}^a r^3 drd\theta$
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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 $\displaystyle \Theta$, 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 $\displaystyle -\infty<x<+\infty$, then when I use $\displaystyle x=rcos\Theta$, what are my equivalent bounds for r and Theta?

    I am thinking $\displaystyle 0<\Theta<2\pi$ and $\displaystyle 0<r<+\infty$.
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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 $\displaystyle -\infty < x< \infty$ and $\displaystyle -\infty< y< \infty$, then that would be the entire xy-plane and, in polar coordinates, $\displaystyle 0\le \theta\le 2\pi$, $\displaystyle 0\le r< \infty$ 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 $\displaystyle -\infty< x< \infty$ and $\displaystyle 0\le y< \infty$, then you would have the upper half plane and the bounds, in polar coordinates, would be $\displaystyle 0\le \theta\le \pi$, $\displaystyle 0\le r< \infty$.
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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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