(Problem 1a) Compute where is the rectangular region bounded by the planes and
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In this section we compute triple integrals over various regions.
A triple integral is computed in a manner similar to a double integral. Instead of integrating over a region in the -plane, we integrate over regions in -space. Thus, triple integrals will become three iterated integrals rather than two. For simplicity, we will assume that our regions include boundaries for the variable of the form and we will integrate either or .
Example 1 Compute where is the rectangular region bounded by the planes and
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The region is rectangular with constant boundaries for each of the three variables. Furthermore, the integrand is a separable function, i.e., . Hence, our iterated integrals can be written as a product of three ordinary definite integrals. We have:
The region is rectangular with constant boundaries for each of the three variables. Furthermore, the integrand is a separable function, i.e., . Hence, our iterated integrals can be written as a product of three ordinary definite integrals. We have:
\begin{align*} \iiint _R xyz \, dV & = \int _0^3\int _1^2 \int _{-2}^3 xyz \,dz\, dy \, dx\\ & =\left (\int _0^3 x \, dx \right ) \cdot \left (\int _1^2 y\, dy\right ) \cdot \left (\int _{-2}^3z \,dz \right )\\ & = \frac 92 \cdot \left (\frac{4}{2} -\frac 12\right ) \cdot \left (\frac 92 - \frac 42\right )\\ &= \frac{135}{8} \end{align*}
Example 2 Compute where is the tetrahedron in the first octant with vertices and
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The plane determined by the points and has normal vector and has equation Hence, the boundaries for come from
The base of the tetrahedron is the triangle in the -plane with vertices and . The line connecting the points and has equation . Thus, the base of the tetrahedron can be described using the inequalities The triple integral over the tetrahedron can now be written as iterated integrals and computed: \begin{align*} \iiint _R x \, dV & = \int _0^1 \int _0^{-2x+2} \int _0^{3 - 3x - 3y/2} x \, dz\, dy \, dx\\ & = \int _0^1 \int _0^{-2x+2} xz\bigg |_0^{3 - 3x - 3y/2} \, dy \, dx\\ & = \int _0^1 \int _0^{-2x+2} (3x - 3x^2 - 3xy/2) \, dy \, dx\\ &= \int _0^1 (3xy - 3x^2y - 3xy^2/4)\bigg |_0^{-2x+2} \, dx\\ &= \int _0^1 3x(-2x+2) -3x^2(-2x+2) -3x(-2x+2)^2/4 \, dx\\ &= \int _0^1 (3x^3 - 6x^2 + 3x) \, dx\\ &= \frac 34 - 2 + \frac 32 = \frac 14 \end{align*}
Here is a video solution of problem 2:
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Example 3 Compute where is the region in the first octant below the surface and
above the triangle in the -plane with vertices and .
The boundaries for are . The boundaries for and come from the triangle bounded by the lines and . We choose to integrate with respect to and then with respect to because the integrand is (slightly) easier to deal with. This detail is not evident until after we integrate with respect to . We have:
\begin{align*} \iiint _R e^{z/y} \, dV & = \int _0^1 \int _y^1 \int _0^{xy} e^{z/y} \, dz \, dx \, dy\\ & = \int _0^1 \int _y^1 ye^{z/y}\bigg |_{0}^{xy} \,dx \, dy\\ & = \int _0^1 \int _y^1 \left (ye^x - y\right ) \,dx \, dy\\ &= \int _0^1 \left (ye^x - xy\right ) \bigg |_y^1 \, dy\\ &= \int _0^1 \left (ye - y - ye^y + y^2\right ) \, dy\\ &= \frac{y^2 e}{2} - \frac{y^2}{2} - (y-1)e^y + \frac{y^3}{3} \bigg |_0^1\\ &= \frac{e}{2} - \frac 12 + \frac 13 - (-1)\\ &= \frac{e-1}{2} - \frac 23 \end{align*}
(Problem 3) Compute where is the region in the first octant below the surface and
above the triangle in the -plane with vertices and .
Here is a video solution of problem 3:
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