Calculate the length of the curve given by x = e2theta costheta and y = e2theta sintheta for 0 theta pi. Calculate the surface area generated when the curve of y = cosx between x = 0 and x = (pi)/(2) rotates about the x-axis.

Given $x = e^{2\theta} \cos\theta$:

$$\frac{dx}{d\theta} = \frac{d}{d\theta}(e^{2\theta} \cos\theta) = (2e^{2\theta})\cos\theta + e^{2\theta}(-\sin\theta) = e^{2\theta}(2\cos\theta - \sin\theta)$$

Given $y = e^{2\theta} \sin\theta$:

$$\frac{dy}{d\theta} = \frac{d}{d\theta}(e^{2\theta} \sin\theta) = (2e^{2\theta})\sin\theta + e^{2\theta}(\cos\theta) = e^{2\theta}(2\sin\theta + \cos\theta)$$

$$\left(\frac{dx}{d\theta}\right)^2 = (e^{2\theta}(2\cos\theta - \sin\theta))^2 = e^{4\theta}(4\cos^2\theta - 4\sin\theta\cos\theta + \sin^2\theta)$$ $$\left(\frac{dy}{d\theta}\right)^2 = (e^{2\theta}(2\sin\theta + \cos\theta))^2 = e^{4\theta}(4\sin^2\theta + 4\sin\theta\cos\theta + \cos^2\theta)$$

Summing these:

$$\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = e^{4\theta}(4\cos^2\theta - 4\sin\theta\cos\theta + \sin^2\theta + 4\sin^2\theta + 4\sin\theta\cos\theta + \cos^2\theta)$$ $$= e^{4\theta}(5\cos^2\theta + 5\sin^2\theta) = 5e^{4\theta}(\cos^2\theta + \sin^2\theta) = 5e^{4\theta}(1) = 5e^{4\theta}$$

The arc length formula for parametric curves is $L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} d\theta$.

$$L = \int_{0}^{\pi} \sqrt{5e^{4\theta}} d\theta = \int_{0}^{\pi} \sqrt{5} \sqrt{e^{4\theta}} d\theta = \int_{0}^{\pi} \sqrt{5} e^{2\theta} d\theta$$

$$L = \sqrt{5} \left[ \frac{1}{2}e^{2\theta} \right]_{0}^{\pi}$$ $$L = \frac{\sqrt{5}}{2} (e^{2\pi} - e^{2(0)})$$ $$L = \frac{\sqrt{5}}{2} (e^{2\pi} - e^0)$$ $$L = \frac{\sqrt{5}}{2} (e^{2\pi} - 1)$$

The length of the curve is $\boxed{\frac{\sqrt{5}}{2} (e^{2\pi} - 1)}$.

Given $y = \cos x$:

$$\frac{dy}{dx} = \frac{d}{dx}(\cos x) = -\sin x$$

The surface area of revolution about the x-axis is $S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$.

$$S = \int_{0}^{\frac{\pi}{2}} 2\pi (\cos x) \sqrt{1 + \sin^2 x} dx$$

Let $u = \sin x$. Then $du = \cos x \, dx$.
When $x = 0$, $u = \sin(0) = 0$.
When $x = \frac{\pi}{2}$, $u = \sin\left(\frac{\pi}{2}\right) = 1$.
The integral becomes:

$$S = 2\pi \int_{0}^{1} \sqrt{1 + u^2} du$$

This is a standard integral of the form $\int \sqrt{a^2 + u^2} du = \frac{u}{2}\sqrt{a^2+u^2} + \frac{a^2}{2}\ln|u+\sqrt{a^2+u^2}|$. Here $a=1$.

$$S = 2\pi \left[ \frac{u}{2}\sqrt{1+u^2} + \frac{1}{2}\ln|u+\sqrt{1+u^2}| \right]_{0}^{1}$$

Evaluate at the upper limit $u=1$:

$$\frac{1}{2}\sqrt{1+1^2} + \frac{1}{2}\ln|1+\sqrt{1+1^2}| = \frac{1}{2}\sqrt{2} + \frac{1}{2}\ln(1+\sqrt{2})$$

Evaluate at the lower limit $u=0$:

$$\frac{0}{2}\sqrt{1+0^2} + \frac{1}{2}\ln|0+\sqrt{1+0^2}| = 0 + \frac{1}{2}\ln(1) = 0$$

Subtract the lower limit value from the upper limit value:

$$S = 2\pi \left( \left(\frac{\sqrt{2}}{2} + \frac{1}{2}\ln(1+\sqrt{2})\right) - 0 \right)$$ $$S = 2\pi \left( \frac{\sqrt{2}}{2} + \frac{1}{2}\ln(1+\sqrt{2}) \right)$$ $$S = \pi\sqrt{2} + \pi\ln(1+\sqrt{2})$$

The surface area generated is $\boxed{\pi\sqrt{2} + \pi\ln(1+\sqrt{2})}$.