Calculate the volume under the elliptic paraboloid z = 2x^2 + 6y^2 and over the rectangle ...

Question:

Calculate the volume under the elliptic paraboloid {eq}z = 2x^2 + 6y^2 {/eq} and over the rectangle {eq}[-2,2] \times [-2,2] {/eq}

Double Integrals:

In much the same way we used a definite integral to find the area of the region below a curve and above the {eq}x {/eq}-axis, we can use a double integral to find the volume below a surface and above the {eq}xy {/eq}-plane. Specifically, we write

{eq}\begin{align*} V &= \iint_R f (x,y)\ dA \end{align*} {/eq}

Answer and Explanation:

We have {eq}R = [-2,2] \times [-2,2] {/eq}, so the volume below the paraboloid is

{eq}\begin{align*} V &= \iint_R 2x^2 + 6y^2\ dA \\ &= \int_{-2}^2 \int_{-2}^2 2x^2 + 6y^2\ dy \ dx \\ &= \int_{-2}^2 \left [ 2x^2y + 2y^3 \right ]_{-2}^2 \ dx \\ &= \int_{-2}^2 2x^2 (2+2) + 2 (2^3 - (-2)^3)\ dx \\ &= \int_{-2}^2 8x^2 + 32\ dx \\ &= \left [ \frac83x^3 + 32x \right ]_{-2}^2 \\ &= \frac83 \left( 2^3 - (-2)^3 \right) + 32 \left( 2+2 \right) \\ &= \frac{512}{3} \\ &\approx 170.6667 \end{align*} {/eq}


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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
401

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