Let E be the solid inside cylinder y^2+z^2=1 and x^2+z^2=1, find Surface Area of E

[jk8985]

Can someone let me know if this is correct and if I showed all steps?

Since the the region of integration inside x^2 + y^2 = 1 (and symmetry), convert to polar coordinates:
2 * 4 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) (r dr dθ)/(1 - r^2 cos^2(θ))^(1/2)
= 8 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) 2r (1 - r^2 cos^2(θ))^(-1/2) dr dθ
= 8 * ∫(θ = 0 to π/2) (-1/cos^2(θ)) 2(1 - r^2 cos^2(θ))^(1/2) {for r = 0 to 1} dθ
= 16 ∫(θ = 0 to π/2) (1/cos^2(θ)) [1 - (1 - cos^2(θ))^(1/2)] dθ
= 16 ∫(θ = 0 to π/2) (1 - sin θ) dθ/cos^2(θ)
= 16 ∫(θ = 0 to π/2) (sec^2(θ) - sec θ tan θ) dθ
= 16 (tan θ - sec θ) {for θ = 0 to π/2}
= 16 (sin θ - 1)/cos θ {for θ = 0 to π/2}
= 16 (0 - (-1)), using L'Hopital's Rule as θ→ π/2-
= 16.

 

[jk8985]

Could you also make it more readable And/or explain it?

 

[stapel]

Is the following re-formatted as desired?

$\displaystyle (2)(4)\displaystyle{\int_{\theta=0}^{\frac{\pi}{2}}\, \int_{r=0}^{1}\, \frac{r}{\sqrt{1\, -\, r^2 \cos^2(\theta)}}\, dr\, d\theta}$

$\displaystyle =\, (8)\displaystyle{\int_{\theta=0}^{\frac{\pi}{2}}\, \int_{r=0}^{1}\,(2r)(1\, -\, r^2 \cos^2(\theta))^{-\frac{1}{2}}\, dr\, d\theta}$

$\displaystyle =\, (8)\displaystyle{\int_{\theta=0}^{\frac{\pi}{2}}\, \left.\left(\frac{-1}{\cos^2(\theta)}\right)\left(2\sqrt{1\, -\, r^2 \cos^2(\theta)}\right)\right|_{r=0}^1\, d\theta}$

$\displaystyle =\, (16)\displaystyle{\int_{\theta=0}^{\frac{\pi}{2}} \,\left(\frac{1}{\cos^2(\theta)}\right)\left(1\, -\, \sqrt{1\, -\, \cos^2(\theta)}\right)\, d\theta}$

$\displaystyle =\, (16)\displaystyle{\int_{\theta=0}^{\frac{\pi}{2}} \,\frac{1\, -\, \sin(\theta)}{\cos^2(\theta)}\, d\theta}$

$\displaystyle =\, (16)\displaystyle{\int_{\theta=0}^{\frac{\pi}{2}} \,\left(\sec^2(\theta)\, -\, \sec(\theta)\tan(\theta)\right)\, d\theta}$

$\displaystyle =\, (16)\displaystyle{\left.\left(\tan(\theta)\, -\, \sec(\theta)\right)\right|_{\theta=0}^{\frac{\pi}{2}}}$

$\displaystyle =\, (16)\displaystyle{\left.\left(\frac{\sin(\theta)\, -\, 1}{\cos(\theta)}\right)\right|_{\theta=0}^{\frac{ \pi }{2}}}$