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The integral that gives the volume of the "shed" with a base R= [(x,y): -1 leq x leq 2, 0 leq y...

Question:

The integral that gives the volume of the "shed" with a base {eq}R= [(x,y): -1 \leq x \leq 2,\ \ 0 \leq y \leq 3] {/eq} and a roof {eq}z = f(x,y) = 18-3x-2y\ {/eq} is?

Using a Triple Integral to Represent a Volume

To calculate the volume of a region {eq}R {/eq} in three dimensions we can use the triple integral {eq}\displaystyle\iiint_R dV. {/eq} If we know the functions {eq}z = f(x, y) {/eq} and {eq}z = g(x, y) {/eq} that define the lower and upper surfaces that bound the region {eq}R, {/eq} then we use these functions for the limits of integration for the variable {eq}z. {/eq} Next we can project the region {eq}R {/eq} onto the {eq}xy {/eq} plane to determine the limits on {eq}x {/eq} and {eq}y. {/eq}

Answer and Explanation:

We are given {eq}R= [(x,y): -1 \leq x \leq 2,\ \ 0 \leq y \leq 3] {/eq} and {eq}z = f(x,y) = 18-3x-2y. {/eq} The floor of our shed is represented by the function {eq}z = 0. {/eq} Therefore a triple integral that represents the volume is

{eq}V = \displaystyle\int_{-1}^2 \int_0^3 \int_0^{18 - 3x - 2y} dz \: dy \: dx. {/eq}


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