# What is the surface area produced by rotating f(x)=sinx-cosx, x in [0,pi/4] around the x-axis?

Jun 2, 2018

${S}_{A} = 2 \pi {\int}_{0}^{\frac{\pi}{4}} \left(\sin x - \cos x\right) \cdot \sqrt{2 + \sin 2 x} \cdot \mathrm{dx} = - 3.7516$

#### Explanation:

The surface area due to $\text{x-axis}$ given by:

color(red)[S_A=2piint_a^by*sqrt(1+(y')^2)*dx

$y = \sin x - \cos x$

$y ' = \cos x + \sin x$

${\left(y '\right)}^{2} = {\cos}^{2} x + 2 \sin x \cdot \cos x + {\sin}^{2} x$

the interval of the integral $x \in \left[0 , \frac{\pi}{4}\right]$

now let setup the interval of the definite integral to determine the surface area:

${S}_{A} = 2 \pi {\int}_{0}^{\frac{\pi}{4}} \left(\sin x - \cos x\right) \cdot \sqrt{1 + {\left(\cos x + \sin x\right)}^{2}} \cdot \mathrm{dx}$

${S}_{A} = 2 \pi {\int}_{0}^{\frac{\pi}{4}} \left(\sin x - \cos x\right) \cdot \sqrt{1 + {\cos}^{2} x + 2 \sin x \cdot \cos x + {\sin}^{2} x} \cdot \mathrm{dx}$

${S}_{A} = 2 \pi {\int}_{0}^{\frac{\pi}{4}} \left(\sin x - \cos x\right) \cdot \sqrt{1 + 1 + \sin 2 x} \cdot \mathrm{dx}$

${S}_{A} = 2 \pi {\int}_{0}^{\frac{\pi}{4}} \left(\sin x - \cos x\right) \cdot \sqrt{2 + \sin 2 x} \cdot \mathrm{dx}$

$= 2 \pi \left[\frac{\sqrt[4]{17} \cdot \sin \left(\arctan \frac{\frac{1}{4}}{2}\right)}{2} ^ \left(\frac{7}{2}\right) + \frac{\sqrt[4]{17} \cdot \cos \left(\arctan \frac{\frac{1}{4}}{2}\right)}{2} ^ \left(\frac{7}{2}\right) - \frac{\sqrt{5}}{8} - \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right]$

$= 2 \pi \left[\frac{8 \sqrt[4]{17} \cdot \sin \left(\arctan \frac{\frac{1}{4}}{2}\right) + 8 \sqrt[4]{17} \cdot \cos \left(\arctan \frac{\frac{1}{4}}{2}\right) - {2}^{\frac{7}{2}} \cdot \sqrt{5} - 64 \sqrt{3} + 64}{2} ^ \left(\frac{13}{2}\right)\right]$

$= - 3.7516$

show below the surface area revolving (shaded):