Answer: (a) The arc length is \(\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\). For \(y=-\ln(\cos x)\),

\(\frac{dy}{dx}=\tan x\), so \(\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\tan^2 x}=\sec x\).

Hence

\(s=\int_0^{\pi/3} \sec x\,dx=[\ln(\sec x+\tan x)]_0^{\pi/3}=\ln(2+\sqrt3).\)

(b) The surface area formed by rotating a curve about the \(x\)-axis is \(S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\).

Here \(y=2\sqrt{x+3}\), so \(\frac{dy}{dx}=\frac{1}{\sqrt{x+3}}\).

Therefore

\(S=2\pi\int_0^1 2\sqrt{x+3}\,\sqrt{1+\frac{1}{x+3}}\,dx=4\pi\int_0^1 \sqrt{x+3}\,\sqrt{\frac{x+4}{x+3}}\,dx.\)

This simplifies to

\(S=4\pi\int_0^1 \sqrt{x+4}\,dx.\)

Now

\(S=4\pi\left[\frac{2}{3}(x+4)^{3/2}\right]_0^1=\frac{8\pi}{3}\left(5^{3/2}-4^{3/2}\right).\)

So

\(S=\frac{8\pi}{3}(5\sqrt5-8).\)

(a)

To find the length of a curve \(y=f(x)\) from \(x=a\) to \(x=b\), use

\(s=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.\)

Given \(y=-\ln(\cos x)\), differentiate:

\(\frac{dy}{dx}=-\frac{-\sin x}{\cos x}=\tan x.\)

So

\(\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\tan^2 x}=\sec x.\)

Hence the arc length from \(x=0\) to \(x=\pi/3\) is

\(s=\int_0^{\pi/3} \sec x\,dx.\)

Using \(\int \sec x\,dx=\ln|\sec x+\tan x|+C\),

\(s=[\ln(\sec x+\tan x)]_0^{\pi/3}.\)

Now \(\sec(\pi/3)=2\) and \(\tan(\pi/3)=\sqrt3\), while at \(x=0\), \(\sec 0=1\) and \(\tan 0=0\). So

\(s=\ln(2+\sqrt3)-\ln(1)=\ln(2+\sqrt3).\)

(b)

For rotation about the \(x\)-axis, the surface area is

\(S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.\)

Here \(y=2\sqrt{x+3}=2(x+3)^{1/2}\), so

\(\frac{dy}{dx}=2\cdot\frac12(x+3)^{-1/2}=\frac{1}{\sqrt{x+3}}.\)

Therefore

\(S=2\pi\int_0^1 2\sqrt{x+3}\,\sqrt{1+\frac{1}{x+3}}\,dx.\)

Simplify the square root:

\(1+\frac{1}{x+3}=\frac{x+4}{x+3},\)

so

\(\sqrt{x+3}\,\sqrt{1+\frac{1}{x+3}}=\sqrt{x+3}\,\sqrt{\frac{x+4}{x+3}}=\sqrt{x+4}.\)

Thus

\(S=4\pi\int_0^1 \sqrt{x+4}\,dx.\)

Integrating,

\(\int \sqrt{x+4}\,dx=\int (x+4)^{1/2}\,dx=\frac{2}{3}(x+4)^{3/2}.\)

So

\(S=4\pi\left[\frac{2}{3}(x+4)^{3/2}\right]_0^1=\frac{8\pi}{3}\left(5^{3/2}-4^{3/2}\right).\)

Now \(5^{3/2}=5\sqrt5\) and \(4^{3/2}=8\), giving

\(S=\frac{8\pi}{3}(5\sqrt5-8).\)