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Compute the following integral $$\iiint_D xydzdydx$$

Where $D$ is the space region restricted to $z=4-x^2-y^2$ and $x^2+y^2=1$ and $z=0$.

Here is a plot:

enter image description here

So I think the triple integral is indeed: $$\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{0}^{4-x^{2}-y^{2}}xy dzdydx$$

But I'm not sure about that, moreover I want to compute the triple integral without using spherical coordinate system.

Stephen Donovan
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masaheb
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  • Spherical coordinates is overkill anyway. Are you familiar with cylindrical coordinates? On the other hand, you can get an immediate result by observing the symmetry of the integrand in the $x,y$ plane... – user170231 Apr 13 '21 at 14:16
  • You don't need cylindrical coordinates to integrate an odd function over the even region. – Vasily Mitch Apr 13 '21 at 14:22

1 Answers1

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Your integral is set up correctly. But the point to note is that $f(x,y) = xy$ is an odd function wrt to $x$ and $y$. In other words, $f(-x,y) = -xy = -f(x,y)$ and $f(x,-y) = - xy = -f(x,y)$.

Given the region is symmetric to $YZ$ plane, the integral for $x \geq 0$ and for $x \leq 0$ will cancel each other out. So the integral is zero.

EDIT: That aside, if you were to set it up in cylindrical coordinates,

$x = r \cos\theta, y = r \sin\theta, x^2+y^2 = r^2 \leq 1$ and bounds are

$0 \leq z \leq 4-r^2, 0 \leq r \leq 1$, so the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^1 \int_0^{4-r^2} r^2 \sin\theta \cos\theta \cdot r \ dz \ dr \ d\theta$

Math Lover
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