Surface Area

Frustum

𝒜 \approx 2 π f (x_{i}^{*}) \sqrt{{(Δ x)}^{2} + {(Δ y)}^{2}}

Let $f (x)$ be a smooth positive function over the interval $[𝒶, 𝒷]$ . Then the surface area formed by revolving the graph of $f (x)$ around the $x$ -axis from $(𝒶, f (𝒶))$ to $(𝒷, f (𝒷))$ is

𝒜 = \int_{𝒶}^{𝒷} 2 π f (x) (\sqrt{1 + {(\frac{ⅆ y}{ⅆ x})}^{2}}) ⅆ x = \int_{𝒶}^{𝒷} 2 π f (x) (\sqrt{1 + {(f' (x))}^{2}}) ⅆ x

Let $g (y)$ be a smooth positive function over the interval $[𝒸, 𝒹]$ . Then the surface area formed by revolving the graph of $g (y)$ around the $y$ -axis from $(𝒸, g (𝒸))$ to $(𝒹, g (𝒹))$ is

𝒜 = \int_{𝒸}^{𝒹} 2 π g (y) (\sqrt{1 + {(\frac{ⅆ x}{ⅆ y})}^{2}}) ⅆ y = \int_{𝒸}^{𝒹} 2 π g (y) (\sqrt{1 + {(g' (y))}^{2}}) ⅆ y

Find the area of the surface of revolution generated by revolving the curve

f (x) = \sqrt{x} 0 \leq x \leq 4

around the $x$ -axis.

Find the area of the surface of revolution generated by revolving the curve

g (y) = \frac{y^{3}}{3} 0 \leq y \leq 2

around the $y$ -axis.