# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 39

$$\frac{384 \pi}{5}$$

#### Work Step by Step

Since $$1+\left(y^{\prime}\right)^{2}=1+\left(-\frac{\sqrt{4-x^{2 / 3}}}{\sqrt[3]{x}}\right)^{2}=1+\frac{4-x^{2 / 3}}{x^{2 / 3}}=\frac{4}{x^{2 / 3}}$$ Then \begin{aligned} \text { Surface Area } &=2 \pi \int_{0}^{8}\left(4-x^{2 / 3}\right)^{3 / 2} \sqrt{\frac{4}{x^{2 / 3}}} d x \\ &=2 \pi \int_{0}^{8}\left(4-x^{2 / 3}\right)^{3 / 2} \frac{2}{x^{1 / 3}} d x \\ &=4 \pi \int_{0}^{8} \frac{\left(4-x^{2 / 3}\right)^{3 / 2}}{\sqrt[3]{x}} d x \end{aligned} Let $$u=4-x^{2 / 3} \longrightarrow d x=-\frac{3 \sqrt[3]{x}}{2}$$ Then \begin{aligned} \int \frac{\left(4-x^{2 / 3}\right)^{3 / 2}}{\sqrt[3]{x}} d x &=-\frac{3}{2} \int \frac{u^{3 / 2} \sqrt[3]{x}}{\sqrt[3]{x}} d u \\ &=-\frac{3}{2} \int u^{3 / 2} d u \\ &=-\frac{3}{2}\left[\frac{2}{5} u^{5 / 2}\right] \\ & =-\frac{3}{5}\left(4-x^{2 / 3}\right)^{5 / 2} \end{aligned} Hence \begin{aligned} \text { Surface Area }&=4 \pi \int_{0}^{8} \frac{\left(4-x^{2 / 3}\right)^{3 / 2}}{\sqrt[3]{x}} d x &=4 \pi\left[-\frac{3}{5}\left(4-x^{2 / 3}\right)^{5 / 2}\right]_{0}^{8} \\ &=4 \pi\left[\frac{96}{5}\right] \\ &=\frac{384 \pi}{5} \end{aligned}

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