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I have to evaluate this integral:

$$ \int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}}\int_{0}^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2+z^2}dzdxdy $$

in spherical coordinates. I see that the region in the xy plane is a circular sector bound by $y=x$ and $y=\sqrt2$ with a radius of 2, I have found that the region in three dimensions becomes complicated to evaluate because of the plane that cuts the spherical sector at y=sqrt(2). I am having trouble finding an expression for $\rho$ or r that describes both the spherical part and the planar part, as well as an $\phi$ that works as well, is see that $ \frac{\pi }{4}\leq \theta \leq \pi $.

Thanks

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I found a solution: I take the integral of the spherical sector, then I subtract out the area cut off by the plane to get the answer. –  Kyle Payne Apr 6 '12 at 16:33
    
Perhaps I'm miscalculating but: In the $x$-$y$ plane you have $0\le y\le\sqrt2$ and $y\le x\le\sqrt{4-y^2}$. This region is a quarter-circle in the first quadrant of the $x$-$y$ plane of radius 2 with one side on the positive $x$-axis. No? –  David Mitra Apr 6 '12 at 16:40
    
@David: I think you mean an eighth circle? –  joriki Apr 6 '12 at 16:47
    
@joriki Yes, indeed; thanks. –  David Mitra Apr 6 '12 at 16:48
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2 Answers 2

As David (almost) pointed out, the bounds of the first two integrals correspond to an eighth circle of radius $2$ with $0\le\phi\le\pi/4$. The bound on $z$ also corresponds to $r=2$, so the integral is just

$$\int_0^2\int_0^{\pi/2}\int_0^{\pi/4}rr^2\sin\theta\mathrm d\phi\mathrm d\theta\mathrm dr=\frac142^4\cdot1\cdot\frac\pi4=\pi\;.$$

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I don't think you have the correct region.

Of course to set up the integral in spherical coordinates, you need to describe the region in spherical coordinates.

So, let's do that. First, from the given integral, let's determine the solid $S$ defined by the limits of integration. Towards this end, let's call the innermost integral $g(x,y)$. Then we can write the given integral as $$\tag{1} \int_0^{\sqrt 2} \int_y^{\sqrt{4-y^2}} g(x,y) \,dx\,dy. $$ We will now determine the region $R$ in the $x$-$y$ plane that the limits of integration in $(1)$ determine. Since the outermost integral is with respect to $y$, the region $R$ is "generated" by line segments $l_y$ that are parallel to the $x$-axis. The line segments start at $y=0$ and end at $y=\sqrt2$ (the limits of integration of the outer integral). For a fixed $y$ in $[0,\sqrt2]$, the endpoints of the line segment $l_y$ are determined by the limits of integration of the inner integral: $l_y$ has endpoints on the graphs of $x=y$ and $x=\sqrt{4-y^2}$.

Alternatively, you can read a description of $R$ from the limits: $$ R=\bigl\{\, (x,y)\mid 0\le y\le \sqrt2, y\le x\le \sqrt{4-y^2}\,\bigr\}. $$

Either way, we see that $R$ is the blueish-shaded region below:

enter image description here


Now back to the triple integral. Considering the inner most integral, we see that the solid is bounded below by the region $R$ (as the lower limit of integration is $z=0$) and bounded above by the sphere of radius 2 (as the upper limit of integration is $z=\sqrt{4-x^2-y^2}$).

So $S$ is a spherical sector and is described by the spherical coordinates:

$$ \eqalign{ 0&\le r\le 2\cr 0&\le\theta\le\pi/4\cr 0&\le \phi\le\pi/2 } $$ (I use $\phi$ as the angle to the $z$-axis.)

And from this you can now set up the integral in spherical coordinates.

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