What is the surface area produced by rotating f(x)=x/pi^2, x in [-3,3] around the x-axis?

May 26, 2018

$\textcolor{b l u e}{{S}_{A} = 2 \pi {\int}_{-} {3}^{3} \frac{x}{\pi} ^ 2 \cdot \sqrt{1 + {\left(\frac{1}{\pi} ^ 2\right)}^{2}} \cdot \mathrm{dx} = = \frac{\sqrt{{\pi}^{2} + 1} \cdot 9}{\pi} ^ 3 - \frac{\sqrt{{\pi}^{2} + 1} \cdot 9}{\pi} ^ 3 = 0}$

Explanation:

If the solid is obtained by rotating the graph of $y = f \left(x\right)$ from $x = a$ to $x = b$ around the $\text{x-axis}$ then the surface area ${S}_{A}$ can be found by the integral

${S}_{A} = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$

${S}_{A} = 2 \pi {\int}_{-} {3}^{3} \frac{x}{\pi} ^ 2 \cdot \sqrt{1 + {\left(\frac{1}{\pi} ^ 2\right)}^{2}} \cdot \mathrm{dx}$

${S}_{A} = 2 {\int}_{-} {3}^{3} \frac{x}{\pi} \cdot \sqrt{1 + \left(\frac{1}{\pi} ^ 4\right)} \cdot \mathrm{dx}$

${S}_{A} = 2 {\int}_{-} {3}^{3} \frac{x}{\pi} \cdot \sqrt{\frac{1 + {\pi}^{2}}{\pi} ^ 4} \cdot \mathrm{dx}$

${S}_{A} = \frac{2}{\pi} \sqrt{\frac{1 + {\pi}^{2}}{\pi} ^ 4} {\int}_{-} {3}^{3} x \cdot \mathrm{dx} = {\left[\frac{\sqrt{{\pi}^{2} + 1} \cdot {x}^{2}}{\pi} ^ 3\right]}_{-} {3}^{3}$

$= \frac{\sqrt{{\pi}^{2} + 1} \cdot 9}{\pi} ^ 3 - \frac{\sqrt{{\pi}^{2} + 1} \cdot 9}{\pi} ^ 3 = 0$

May 26, 2018

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