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I'm reviewing for my Calculus 3 midterm, and one of the practice problems I'm going over asks to find the volume of the below solid 1. by using a triple integral with spherical coordinates, and 2. by using a triple integral with cylindrical coordinates.

I'm able to do the integral with spherical coordinates, but I'm getting confused on the one with cylindrical coordinates. In my notes I have written that the cylindrical volume should be: $$dv = r\ dr\ d\theta\ dz$$ Looking at the solution to this problem, it is integrated in the order: $dz\ dr\ d \theta$ , I'm not sure what the point is of changing the integration order? Also, the bounds in the solution for the integral with respect to $z$ are from $0$ to $\sqrt{4-r^2}$, and I'm not sure where that is coming from either?

Any help would be greatly appreciated.

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1 Answer 1

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You know the equation of such part of the sphere is $$z^2=4-(x^2+y^2),x\in[0..2],y\in[0..2]$$ But $r^2=x^2+y^2$ and then $z=\sqrt{4-r^2}$. The ranges of our new variables are : $$\theta|_0^{\pi/2}, r|_0^2, z|_0^{\sqrt{4-r^2}}$$ So we have to evaluate $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{4-r^2}}dv$$

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  • $\begingroup$ Nice! and quick...$\quad+1\quad$;-) $\endgroup$
    – amWhy
    Commented Mar 6, 2013 at 6:40
  • $\begingroup$ We haven't learned Jacobian determinant yet. $\endgroup$ Commented Mar 6, 2013 at 6:40
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    $\begingroup$ There are some relations between coordinates. Two of them are $x=r\cos t, y=r\sin t$ which lead us to $x^2+y^2=r^2$ relating the Cartesian and Cylindrical ones. $\endgroup$
    – Mikasa
    Commented Mar 6, 2013 at 7:09
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    $\begingroup$ @PseudoPsyche: They come from when you want to find the coordinates $(x,y)$ of an arbitrary point lying on a circle with radius $r$ centered at the origin. :-) $\endgroup$
    – Mikasa
    Commented Mar 6, 2013 at 7:35
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    $\begingroup$ Alright, thanks! This is the last section we covered before the exam, and it's the only thing I'm really iffy on so overall I don't think it'll be too bad. $\endgroup$ Commented Mar 6, 2013 at 8:04

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