Solutions - 0090-03

Method 1: integrate in $x$

(1) Integral formula for surface area:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+{(f^{\prime })}^2}dx\gg \gg 2\pi {\int }_0^{2}x\sqrt{1+4x^2}dx$$

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(2) Integrate: perform $u$-sub with $u=1+4x^2$ and so $du=8xdx$:

$$\begin{array}{c}S\gg \gg \frac{\pi }{4}{\int }_1^{17}\sqrt{u}du\\ \\ \gg \gg \frac{\pi }{4}{\frac{2}{3}{u}^{\frac{3}{2}}|}_1^{17}\\ \\ \gg \gg \frac{\pi }{6}({17}^{3/2}-1)\end{array}$$

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Method 2: integrate in $y$

(1) Integral formula for surface area using $g(y)=\sqrt{y}$:

$$\begin{array}{c}S={\int }_c^{d}2\pi g(y)\sqrt{1+{(g^{\prime })}^2}dy\gg \gg {\int }_0^{4}2\pi \sqrt{y}\sqrt{1+\frac{1}{4}{y}^{-1}}dy\\ \\ \gg \gg \pi {\int }_0^4\sqrt{4y+1}dy\end{array}$$

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(2) Integrate: perform $u$-sub with $u=4y+1$ and so $du=4dy$:

$$\begin{array}{c}\pi {\int }_0^4\sqrt{4y+1}dy\gg \gg \frac{\pi }{4}{\int }_1^{17}\sqrt{u}du\\ \\ \gg \gg \frac{\pi }{4}{\frac{2}{3}{u}^{\frac{3}{2}}|}_1^{17}\\ \\ \gg \gg \frac{\pi }{6}({17}^{3/2}-1)\end{array}$$