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Find the volume of the solid bounded above by the surface $z=f(x, y)$ and below by the region $R$ in the $x$ -y plane. $f(x, y)=x, R$ : rectangle determined by $0 \leq x \leq 1,0 \leq y \leq 2$.

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Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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For this problem, we are asked to find the volume of the solid bounded above by the surface. Z equals X and below by the region, or by the rectangle determined by X between zero and one and why between zero and two in the Xy plane. So we'll be finding the integral from, let's see here, we can do this as the integral from 0 to 2 of the integral from 0 to 1 of X. Dx Dy. So this will then be the interval from 0 to 2 of X squared over two, evaluated from 0 to 1. Dy. So that will then be 1/2 times the integral from 0 to 2 of just Dy. So this would be 1/2 times two, Giving us that the volume will be equal to one.

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