Volume Solution 3

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SOLUTION 3: a.) Here are sketches of the base of the solid and the entire solid.

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Here are sketches of a square cross-section at

$x$ , together with it's dimensions.

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The area of the square cross-section is

$A(x)= (edge)^2 = (8-x^3)^2$ . Thus the total volume of this static solid is

$Volume = \int_{0}^{2} (8-x^3)^2 \ dx$

SOLUTION 3: b.) Here are sketches of the base of the solid and the entire solid.

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Here are sketches of an equilateral triangular cross-section at

$x$ , together with its dimensions. We need to figure out the height and area of this equilateral triangle.

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Consider an equilateral triangle of edge length

$z$ and height

$h$ in the diagram below.

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Using the Pythagorean Theorem we get

$h^2 + \Big({ z \over 2 }\Big)^2=z^2 \ \ \ \ \longrightarrow$

$h^2 + { z^2 \over 4 }=z^2 \ \ \ \ \longrightarrow$

$h^2 = { 3 \over 4 }z^2 \ \ \ \ \longrightarrow$

$h = { \sqrt{3} \over 2 }z$
Thus the area of an equilateral triangle with edge

$z$ is

$Area = {1 \over 2} (base)(height) = {1 \over 2}(z)\Big({ \sqrt{3} \over 2 }z \Big) = { \sqrt{3} \over 4 }z^2$
so that the area of the equilateral triangular cross-section of edge length

$8-x^3$ is

$A(x)= { \sqrt{3} \over 4 }(8-x^3)^2$ . Thus the total volume of this static solid is

$Volume = \int_{0}^{2} { \sqrt{3} \over 4 }(8-x^3)^2 \ dx$

SOLUTION 3: c.) Here are sketches of the base of the solid and the entire solid.

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Here are sketches of a rectangular cross-section at

$x$ , together with its dimensions, where the height

$h$ is unknown.

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To find the height

$h$ we use the triangles's perimeter of 16 getting

$2h + 2(8-x^3) = 16 \ \ \ \ \longrightarrow$

$h + (8-x^3) = 8 \ \ \ \ \longrightarrow$

$h = x^3$
so that the area of the rectangular cross-section is

$A(x)= (height)(base) = (x^3)(8-x^3)= 8x^3-x^6$ . Thus the total volume of this static solid is

$Volume = \int_{0}^{2} (8x^3-x^6) \ dx$

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