To find the surface area of a revolution, we must use the following formula.

\(\displaystyle SA= 2\pi \int_{a}^{b} r \cdot ds\), where "r" is the variable radius of the curve from the axis of revolution, and "ds" is the differential of the arc length of the curve at that radius. For this problem, "r" is simply the variable "x", because every point's radius from the axis of revolution (the y axis), is its x coordinate.

You can think of "ds" as the arc length formula after it is differentiated. Arc length is an integral, so the derivative of an integral is just what's inside that integral. Here that means \(\displaystyle ds=\sqrt{1+(f'(x))^{2}} \cdot dx\).

"a" and "b" will be the x=0 and x=2 part of the question.

From this information we can rewrite the surface area formula as

\(\displaystyle SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +(f'(x))^2} \cdot dx\)

Now we will find f'(x).

\(\displaystyle f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 1)=2x\)

Plug \(\displaystyle 2x\) in to the formula.

\(\displaystyle SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +(2x)^2} \cdot dx\)

Now simplify \(\displaystyle (2x)^2\).

\(\displaystyle SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +4x^2} \cdot dx\)

Notice that the \(\displaystyle x\) is part of the derivative of what is under the radical. This almost follows the basic integral form \(\displaystyle \int{u^n \cdot du}\),with \(\displaystyle u= 1 + 4x^2\), and \(\displaystyle du = 8x\). Let's force it to match the form exactly. First I will rewrite the square root as an exponent.

\(\displaystyle SA = 2 \pi \int_{0}^{2} x \cdot (1 +4x^2)^{\frac{1}{2}} \cdot dx\)

Now multiply by a form of 1 (highlighted in red) to force the "du" part of the integration form.

\(\displaystyle SA = 2 \pi \cdot {\color{Red} \frac{1}{8}} \int_{0}^{2} {\color{Red} 8} \cdot x \cdot (1 +4x^2)^{\frac{1}{2}} \cdot dx\)

It may help to rewrite the expression in terms of \(\displaystyle u\) to make it easier to apply integration rules. We just need to remember to convert back to x after integrating, before we plug in the bounds.

\(\displaystyle 2\pi \cdot \frac{1}{8} \int u^\frac{1}{2} du\)

Now integrate.

\(\displaystyle \frac{\pi}{4} \cdot [\frac{2}{3}u^{\frac{3}{2}}]\)

Now convert u back to x. Recall that \(\displaystyle u= 1 + 4x^2\)

\(\displaystyle SA = \frac{\pi}{4} \cdot [\frac{2(1 + 4x^2)^{\frac{3}{2}}}{3}]|_{0}^{2}\)

Now plug in the upper and lower bounds.

\(\displaystyle SA = \frac{\pi}{4} \cdot [(\frac{2(1 +4(2)^2)^{\frac{3}{2}}}{3})-(\frac{2(1 + 4(0)^2)^{\frac{3}{2}}}{3})]\)

Now use a calculator to find the approximate answer of

\(\displaystyle SA \approx 36.18\)

To solve for the average value on the interval \(\displaystyle [a,b]\) we follow the following formula

\(\displaystyle \frac{\int_{a}^{b}f(x)\,dx}{b-a}\)

For this problem we evaluate

\(\displaystyle \frac{\int_{0}^{10}2x\,dx}{10-0}\)

\(\displaystyle =\frac{x^2|_0^{10}}{10}\)

\(\displaystyle =\frac{10^2-0^2}{10}\)

\(\displaystyle =\frac{100}{10}\)

\(\displaystyle =10\)

To solve for the average value on the interval \(\displaystyle [a,b]\) we follow the following formula

\(\displaystyle \frac{\int_{a}^{b}f(x)\,dx}{b-a}\)

For this problem we evaluate

\(\displaystyle \frac{\int_{0}^{3}3x^2\,dx}{3-0}\)

\(\displaystyle =\frac{x^3|_0^{3}}{3}\)

\(\displaystyle =\frac{3^3-0^3}{3}\)

\(\displaystyle =\frac{27}{3}\)

\(\displaystyle =9\)

Recall the formula for average value of a function, f(x), on a closed interval, [a,b], is: \(\displaystyle f(avg)=\frac{1}{b-a}\int_{a}^{b}f(x)dx\) 

For this we first need to integrate our function on the defined interval:

\(\displaystyle \int_{0}^{3}(x^3-2x^2+4)dx=(\frac{x^4}{4}-\frac{2x^3}{3}+4x)_{x=0}^{x=3}=\frac{57}{4}\)

Plugging this into our formula we attain

\(\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{3}\cdot \frac{57}{4}=\frac{19}{4}\)

To find the average value, we need to evaluate \(\displaystyle \frac{1}{b-a}\int_a^bf(x)dx =\frac{1}{\pi/2-0}\int_0^{\pi/2}\frac{3x}{1+x^4}dx\).

This integral can be evaluated using \(\displaystyle u\)-substitution.

\(\displaystyle u = x^2\)

\(\displaystyle du=2x dx\)

Then we can proceed as follows

\(\displaystyle \int \frac{3x}{1+x^4}dx\) (Start)

\(\displaystyle = \frac{3}{2}\int \frac{2x}{1+x^4}dx\) (Factor out a \(\displaystyle 3/2\), leaving a \(\displaystyle 2x\) in the numerator)

\(\displaystyle =\frac{3}{2}\int \frac{1}{1+u^2}du\) (Substitute the equations for \(\displaystyle u, du\))

\(\displaystyle =\frac{3}{2}\tan^{-1}(u)+C\) (Integrate, recall that \(\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}\))

\(\displaystyle =\frac{3}{2}\tan^{-1}(x^2)+C\) (Substitute \(\displaystyle x^2\) back in)

Now we have

\(\displaystyle \frac{1}{b-a}\int_a^bf(x)dx =\frac{1}{\pi/2-0}\int_0^{\pi/2}\frac{3x}{1+x^4}dx = \frac{2}{\pi}[\frac{3}{2}\tan^{-1}(x^2)]_0^{\pi/2}\)

\(\displaystyle \frac{3}{\pi}\tan^{-1}\frac{\pi^2}{4}\).

To solve for the average value on the interval \(\displaystyle [a,b]\) we follow the following formula

\(\displaystyle \frac{\int_a^bf(x)\,dx}{b-a}\)

For this problem we evaluate

\(\displaystyle \frac{\int_0^{10}2x+4\,dx}{10-0}\)

\(\displaystyle =\frac{x^2+4x|_0^{10}}{10}\)

\(\displaystyle =\frac{10^2+4(10)-[0^2+4(0)]}{10}\)

\(\displaystyle =14\)

In order to find the average value we must solve

\(\displaystyle \frac{1}{e-1}\int_{1}^{e} \frac{1}{x}dx=\frac{1}{e-1}ln(x)\) evaluated between 1 and e.

\(\displaystyle =\frac{1}{e-1}(1-0)=\frac{1}{e-1}\)

To solve for the average value on the interval \(\displaystyle [a,b]\) we follow the following formula

\(\displaystyle \frac{\int_a^b\,f(x)\,dx}{b-a}\)

For this problem we evaluate

\(\displaystyle \frac{\int_0^{10}\,\frac{1}{5}x\,dx}{10-0}\)

\(\displaystyle =\frac{\frac{1}{10}x^2|_0^{10}}{10}\)

\(\displaystyle =\frac{\frac{1}{10}(10)^2-\frac{1}{10}(0)^2}{10}\)

\(\displaystyle =\frac{10}{10}\)

\(\displaystyle =1\)

To solve for the average value on the interval \(\displaystyle [a,b]\) we follow the following formula

\(\displaystyle \frac{\int_a^b\,f(x)\,dx}{b-a}\)

For this problem we evaluate

\(\displaystyle \frac{\int_0^{2}\,4x\,dx}{2-0}\)

\(\displaystyle =\frac{2x^2|_0^{2}}{2}\)

\(\displaystyle =\frac{2(2)^2-2(0)^2}{2}\)

\(\displaystyle =\frac{8}{2}\)

\(\displaystyle =4\)