Evaluate the triple integral \iiint_E x^2 e^y \, dV where E is the region bounded by the...

Question:

Evaluate the triple integral {eq}\displaystyle\iiint_E x^2 e^y \, dV {/eq} where {eq}E {/eq} is the region bounded by the parabolic cylinder {eq}z=36 - y^2 {/eq} and the planes {eq}z=0,x=6 \enspace \mathrm{and} \enspace x=-6. {/eq}

Triple Integral in Rectangular Coordinates

In order to evaluate a triple integral of the form {eq}\displaystyle\iiint_{R} f(x, y, z) \, \mathrm{d}V {/eq} we need to determine the limits of integration. If the region {eq}R {/eq} is bounded by graphs of surfaces defined by functions {eq}z = g(x, y) \: \mathrm{and} \: z = h(x, y) {/eq} then we use these functions as limits of integration for {eq}z. {/eq} Then determine the projection of the region {eq}R {/eq} onto the {eq}xy {/eq}-plane and find the appropriate limits for {eq}x \: \mathrm{and} \: y. {/eq}

Answer and Explanation:

We are given {eq}\displaystyle\iiint_E x^2 e^y \, dV {/eq} where {eq}E {/eq} is the region bounded by the parabolic cylinder {eq}z = 36 - y^2 {/eq} and the planes {eq}z=0, \: x=6 \enspace \mathrm{and} \enspace x=-6. {/eq} The cylinder {eq}z = 36 - y^2 {/eq} intersects the {eq}xy {/eq} plane when {eq}z = 0, {/eq} so {eq}36 - y^2 = 0 \\ y = \pm 6 {/eq} Therefore the triple integral becomes

{eq}\begin{eqnarray*}\iiint_E x^2 e^y \, dV & = & \int_{-6}^6 \int_{-6}^6 \int_0^{36 - y^2} x^2 e^y \: dz \: dy \: dx \\ & = & \int_{-6}^6 \int_{-6}^6 \left( x^2 e^y z \right|_0^{36 - y^2} \: dy \: dx \\ & = & \int_{-6}^6 \int_{-6}^6 x^2 e^y (36 - y^2) \: dy \: dx \end{eqnarray*} {/eq}

In order to evaluate {eq}\displaystyle\int e^y( 36 - y^2) \: dy, {/eq} we use integration by parts. Let {eq}u = 36 - y^2, \: dv = e^y \: dy. {/eq} Then {eq}du = -2y \: dy, \: v = e^y {/eq} so

{eq}\displaystyle\int e^y(36 - y^2) \: dy = e^y(36 - y^2) + \int 2y e^y \: dy. {/eq}

We need to integrate by parts again. This time {eq}u = 2y, \: dv = e^y \: dy {/eq} so {eq}du = 2 \: dy, \: v = e^y {/eq} and

{eq}\begin{eqnarray*}\int e^y(36 - y^2) \: dy & = & e^y(36 - y^2) + 2y e^y - \int 2 e^y \: dy \\ & = & e^y(36 - y^2) + 2y e^y - 2e^y + C \end{eqnarray*} {/eq}

Therefore

{eq}\begin{eqnarray*}\int_{-6}^6 \int_{-6}^6 x^2 e^y (36 - y^2) \: dy \: dx & = & \int_{-6}^6 \left( x^2 e^y (36 - y^2 + 2y - 2) \right|_{-6}^6 \: dx \\ & = & \int_{-6}^6 x^2 \left(10 e^6 + 14 e^{-6} \right) \: dx \\ & = & (10 e^6 + 14e^{-6}) \left( \displaystyle\frac{x^3}3 \right|_{-6}^6 \\ & = & 1440 e^6 + 2016 e^{-6} \end{eqnarray*} {/eq}


Learn more about this topic:

Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
15K

Related to this Question

Explore our homework questions and answer library