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Using triple integrals, develop a formula for the volume of a cube of side length s located in...

Question:

Using triple integrals, develop a formula for the volume of a cube of side length{eq}\displaystyle \ s {/eq} located in the 1st octant whose vertices are{eq}\displaystyle \ (0,0,s), \ (s,0,s), \ (s,s,s), \ (0,s,s), \ (s,0,0), \ (0,0,0), \ (0,s,0), and (s,s,0). {/eq}

Finding Volume using a Triple Integral:

If we have a three dimensional solid bounded by {eq}a \leq x \leq b, \: c \leq y \leq d, \: e \leq z \leq f, {/eq} then the volume of the region is given by the triple integral {eq}V = \displaystyle\int_a^b \int_c^d \int_e^f dz \: dy \: dx = (f - e)(d - c)(b - a). {/eq}

Answer and Explanation:

The boundaries of the cube are {eq}0 \leq x \leq s, \: 0 \leq y \leq s, \: 0 \leq z \leq s. {/eq} Therefore the volume of the cube is

{eq}\begin{eqn...

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Volumes of Shapes: Definition & Examples

from GMAT Prep: Tutoring Solution

Chapter 11 / Lesson 9
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