   Chapter 15.1, Problem 38E

Chapter
Section
Textbook Problem

Find the volume of the solid that lies under the hyperbolic paraboloid z = 3y2 − x2 + 2 and above the rectangle R = [−1, 1] × [1, 2].

To determine

To find: The volume of the solid that lies under the hyperbolic paraboloid and above the rectangular region.

Explanation

Calculation:

Formula used:

The volume of the solid, V=RzdA , where, z is the given function.

Given:

The hyperbolic paraboloid is z=3y2x2+2 .

The rectangular region is, R=[1,1]×[1,2] .

Calculation:

The volume of the solid is,

V=RzdA=1112(3y2x2+2)dydx

First, compute the integral with respect to y.

V=11[3y33x2y+2y]12dx=11[y3x2y+2y]12dx

Apply the limit value for y,

V=11[(232x2+2

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