Question

By graphing the boundaries of a 2d figure on a graph using these inequalities{3x-2y>-12,y<3,y>-3, and x<0, what is the volume and surface area of the 3d shape produced by revolving this 2d shape around the y axis?

Answer 1

Question #77546, Math / Geometry

By graphing the boundaries of a 2d figure on a graph using these inequalities $\{3x - 2y > -12, y < 3, y > -3, x < 0\}$ , what is the volume and surface area of the 3d shape produced by revolving this 2d shape around the $y$ -axis?

SOLUTION

A figure bounded by inequalities is shaded on the picture. After revolving this shape around the $y$ -axis we obtain the frustum of a cone (see picture).

To determine the volume we calculate volumes of two cones (big with height 9 and radius 6 and small with height 3 and radius 3) and their subtraction is the required volume:

$$V _ {1} = \frac {1}{3} \pi \cdot 6 ^ {2} \cdot 9 = 1 0 8 \pi ;$$ $$V _ {2} = \frac {1}{3} \pi \cdot 3 ^ {2} \cdot 3 = 9 \pi ;$$ $$V = V _ {1} - V _ {2} = 1 0 8 \pi - 9 \pi = 9 9 \pi .$$

To determine the surface area we calculate area of two cones (big with height 9 and small with height 3) and their subtraction is the required surface area:

$$A _ {1} = \pi \cdot 6 \cdot \sqrt {6 ^ {2} + 9 ^ {2}} = 6 \sqrt {1 1 7} \pi = 1 8 \sqrt {1 3} \pi ;$$ $$A _ {2} = \pi \cdot 3 \cdot \sqrt {3 ^ {2} + 3 ^ {2}} = 3 \sqrt {1 8} \pi = 6 \sqrt {2} \pi ;$$ $$A = A _ {1} - A _ {2} = 1 8 \sqrt {1 3} \pi - 6 \sqrt {2} \pi = 6 (3 \sqrt {1 3} - \sqrt {2}) \pi .$$

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