# Question cb563

Jun 10, 2017

I got about $17.25$, and Wolfram Alpha also does.

The surface area is given by:

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

With $f \left(x\right) = \sin x$, $f ' \left(x\right) = \cos x$, so

$S = 2 \pi {\int}_{0}^{4} \sin x \sqrt{1 + {\cos}^{2} x} \mathrm{dx}$.

But since the graph passes through the $x$ axis at $x = 0 , \pi$ in $\left[0 , 4\right]$...

graph{y = sinx * sqrt(1 + (cosx)^2) [-0.396, 4.47, -1.087, 1.347]}

...we'll split this integral into two parts.

$S = {S}_{1} + {S}_{2}$

$= 2 \pi {\int}_{0}^{\pi} \sin x \sqrt{1 + {\cos}^{2} x} \mathrm{dx} + 2 \pi {\int}_{\pi}^{4} \sin x \sqrt{1 + {\cos}^{2} x} \mathrm{dx}$

Let:

$u = \cos x$
$\mathrm{du} = - \sin x \mathrm{dx}$

This gives:

${S}_{i} = - 2 \pi \int \sqrt{1 + {u}^{2}} \mathrm{du}$

One more substitution. This is of the form $\sqrt{{a}^{2} + {u}^{2}}$, so let:

$u = \tan \theta$
$\mathrm{du} = {\sec}^{2} \theta d \theta$
$\sqrt{1 + {u}^{2}} = \sqrt{1 + {\tan}^{2} \theta} = \sec \theta$

Therefore:

${S}_{i} = - 2 \pi \int {\sec}^{3} \theta d \theta$

If you do not know this integral, I have it here:

$\textcolor{g r e e n}{\int {\sec}^{3} t \mathrm{dt}}$

Let:

$u = \sec t$
$\mathrm{dv} = {\sec}^{2} t \mathrm{dt}$
$\mathrm{du} = \sec t \tan t \mathrm{dt}$
$v = \tan t$

$= \sec t \tan t - \int \sec t {\tan}^{2} t \mathrm{dt}$

$= \sec t \tan t - \int \sec t \left({\sec}^{2} t - 1\right) \mathrm{dx}$

$= \sec t \tan t - \int {\sec}^{3} t \mathrm{dt} + \int \sec t \mathrm{dt}$

$2 \int {\sec}^{3} t \mathrm{dt} = \sec t \tan t + \int \sec t \mathrm{dt}$

$\int {\sec}^{3} t \mathrm{dt} = \frac{1}{2} \left(\sec t \tan t + \ln | \sec t + \tan t |\right) + C$

So, what we have is:

${S}_{i} = - \cancel{2} \pi \left[\frac{1}{\cancel{2}} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right)\right]$

Back-substitution gives:

$= - \pi \left[\sqrt{1 + {u}^{2}} \cdot u + \ln | \sqrt{1 + {u}^{2}} + u |\right]$

$= - \pi \cos x \sqrt{1 + {\cos}^{2} x} - \pi \ln | \sqrt{1 + {\cos}^{2} x} + \cos x |$

Evaluating from $x = 0$ to $\pi$:

$\textcolor{g r e e n}{{S}_{1}} = | \left[- \pi \cos x \sqrt{1 + {\cos}^{2} x} - \pi \ln | \sqrt{1 + {\cos}^{2} x} + \cos x |\right] {|}_{0}^{\pi}$

$= \left[- \pi \cos \pi \sqrt{1 + {\cos}^{2} \pi} - \pi \ln | \sqrt{1 + {\cos}^{2} \pi} + \cos \pi |\right] - \left[- \pi \cos 0 \sqrt{1 + {\cos}^{2} 0} - \pi \ln | \sqrt{1 + {\cos}^{2} 0} + \cos 0 |\right]$

$= \left[\pi \sqrt{2} - \pi \ln | \sqrt{2} - 1 |\right] - \left[- \pi \sqrt{2} - \pi \ln | \sqrt{2} + 1 |\right]$

$= 2 \pi \sqrt{2} - \pi \ln | \sqrt{2} - 1 | + \pi \ln | \sqrt{2} + 1 |$

= color(green)(2pi sqrt(2) - piln|(sqrt(2) - 1)/(sqrt(2) + 1)|#

This is approximately $14.44$.

Finally, evaluating from $x = \pi$ to $x = 4$:

$\textcolor{g r e e n}{{S}_{2}} = | \left[- \pi \cos x \sqrt{1 + {\cos}^{2} x} - \pi \ln | \sqrt{1 + {\cos}^{2} x} + \cos x |\right] {|}_{\pi}^{4}$

$= \left[- \pi \cos 4 \sqrt{1 + {\cos}^{2} 4} - \pi \ln | \sqrt{1 + {\cos}^{2} 4} + \cos 4 |\right] - \left[- \pi \cos \pi \sqrt{1 + {\cos}^{2} \pi} - \pi \ln | \sqrt{1 + {\cos}^{2} \pi} + \cos \pi |\right]$

$= \left[- \pi \cos 4 \sqrt{1 + {\cos}^{2} 4} - \pi \ln | \sqrt{1 + {\cos}^{2} 4} + \cos 4 |\right] - \left[\pi \sqrt{2} - \pi \ln | \sqrt{2} - 1 |\right]$

$= - \pi \cos 4 \sqrt{1 + {\cos}^{2} 4} - \pi \sqrt{2} - \pi \ln | \sqrt{1 + {\cos}^{2} 4} + \cos 4 | + \pi \ln | \sqrt{2} - 1 |$

$= \textcolor{g r e e n}{- \pi \left(\cos 4 \sqrt{1 + {\cos}^{2} 4} + \sqrt{2}\right) - \pi \ln | \frac{\sqrt{1 + {\cos}^{2} 4} + \cos 4}{\sqrt{2} - 1} |}$

This value is around $- 2.83$, but surface area is nonnegative, so we take the absolute value to get $2.83$.

Therefore, the total surface area should be:

$\textcolor{b l u e}{S} = 2 \pi \sqrt{2} - \pi \ln | \frac{\sqrt{2} - 1}{\sqrt{2} + 1} | + \pi \left(\cos 4 \sqrt{1 + {\cos}^{2} 4} + \sqrt{2}\right) + \pi \ln | \frac{\sqrt{1 + {\cos}^{2} 4} + \cos 4}{\sqrt{2} - 1} |$

or about $\textcolor{b l u e}{17.25}$.