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The problem in question is: "Find the volume of the region bounded by $x^2+2y^2 = 2, z = 0$ and $x+y+2z = 2$." Am I setting up the integral correctly?

This is equivalent to integrating the function $z = f(x,y) = \frac{2-x-y}{2}$ over $D = \{ (x,y) | x^2+2y^2 = 2\}$. $D$ can be expressed by the relations $$ - \sqrt{2 - 2y^2} \leq x \leq \sqrt{2 - 2y^2} $$ $$ -1 \leq y \leq 1 $$ Then the volume $V$ is given by $$ V = \frac{1}{2} \iint \limits_D (2 - x - y) dA = \frac{1}{2} \int_{-1}^{1} \int_{ - \sqrt{2 - 2y^2}}^{\sqrt{2-2y^2}} (2-x-y) dx dy $$

Also, when does one use triple integration as opposed to double integration?

My preliminary understanding: in general, it seems like we would use triple integration when the base of the region is not in the plane (for example, when finding the volume of the unit ball). In this example, since $D$ is in the $xy$ plane, double integration suffices.

Ozera
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  • Double vs triple integral is just a matter of preference, most of the time. They just wanted you to practice doing it with a double integral. – apnorton Oct 29 '13 at 23:19
  • To repeat anorton's sentiments: my general rule of thumb is to try to set up a double-integral. If the constraints get too wacky, then try a triple integral. – Eric Stucky Oct 29 '13 at 23:22
  • Thanks for the input. If double vs. triple is a matter of preference, how would one do the above problem with a triple integral?

    How would one compute an integral like $$ \iiint \limits_{B} dx dy dz = \frac{4 \pi}{3} $$ where $B$ denotes the unit ball with a double integral?

     – Ozera Oct 29 '13 at 23:37

1 Answers1

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The integral is set up correctly. As a triple integral, you would write $$ \int_{-1}^1\int_{-\sqrt{2-2y^2}}^{\sqrt{2-2y^2}}\int_0^{\frac{2-x-y}2}\,dz\,dx\,dy. $$

Regarding your question in the comments, if $B$ is the unit ball and $D$ the unit disk on the $xy$-plane, then $$ \iiint_B\,dxdydz=\iint_D2\sqrt{1-x^2-y^2}\,dz\,dydx=2\int_0^{2\pi}\int_0^1\,r\sqrt{1-r^2}\,drdt=\left.-2\pi\,\frac23\,(1-r^2)^{3/2}\right|_0^1=\frac{4\pi}3. $$

Martin Argerami
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