Answer
[$\frac{9\sqrt5}{16}$-$\frac{1}{32}$ln(2+$\sqrt5$)]$\pi$
Work Step by Step
Step 1: The formula to find the surface area is
A=$\int^1_{0}$2$\pi$f(x)$\sqrt {1+(\frac{dy}{dx})^{2}}$dx
The function f(x)=$x^{2}$, so $\frac{dy}{dx}$=2x. A=$\int^1_{0}$2$\pi$$x^{2}$$\sqrt {1+(2x)^{2}}$dx
Step 2: To evaluate this integral, let's use the trigonometry substitution method using the trigonometry relationship shown by the image below. Let's replace $x^{2}$ with $\frac{tan^{2}\theta}{4}$ as x=$\frac{tan\theta}{2}$, replace $\sqrt {1+(2x)^{2}}$ with sec$\theta$, and replace dx with $\frac{1}{2}$$sec^{2}\theta$d$\theta$
So, A=$\int_{}$2$\pi$($\frac{tan^{2}\theta}{4}$)(sec$\theta)$($\frac{1}{2}$$sec^{2}\theta$d$\theta$)=$\frac{1}{4}$$\int$$\pi$$tan^{2}\theta$$sec^{3}\theta$d$\theta$
Now, we use the identity $tan^{2}\theta$=$sec^{2}\theta$-1, so A=$\frac{1}{4}$$\int$$\pi$($sec^{2}\theta$-1)$sec^{3}\theta$d$\theta$
Then, we split the integral, A=$\frac{\pi}{4}$($\int$$sec^{5}\theta$d$\theta$-$\int$$sec^{3}\theta$d$\theta$)
Step 3: To evaluate $\int$$sec^{3}\theta$d$\theta$, first change $\int$$sec^{3}\theta$d$\theta$ into $\int$($sec\theta$)$sec^{2}\theta$d$\theta$.
Then substitute $sec\theta$ with u and substitute $sec^{2}\theta$d$\theta$ with dv. So v= $\int$$sec^{2}\theta$d$\theta$=tan$\theta$
$\int$$sec^{3}\theta$d$\theta$=(sec$\theta$)(tan$\theta$)-$\int$$tan^{2}\theta$$sec\theta$d$\theta$
Use the identity $tan^{2}\theta$=$sec^{2}\theta$-1, so $\int$$sec^{3}\theta$d$\theta$=(sec$\theta$)(tan$\theta$)-$\int$($sec^{2}\theta$-1)$sec\theta$d$\theta$
Then split the integral $\int$$sec^{3}\theta$d$\theta$=(sec$\theta$)(tan$\theta$)-$\int$$sec^{3}\theta$$d\theta$+$\int$$sec\theta$$d\theta$
Solving this equation about $\int$$sec^{3}\theta$$d\theta$, we have:
$\int$$sec^{3}\theta$$d\theta$=$\frac{1}{2}$(sec$\theta$·tan$\theta$+$\int$$sec\theta$$d\theta$)=$\frac{1}{2}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|)
Step 4: To evaluate $\int$$sec^{5}\theta$d$\theta$, first change $\int$$sec^{5}\theta$d$\theta$ into $\int$($sec^{3}\theta$)$sec^{2}\theta$d$\theta$.
Then substitute $sec^{3}\theta$ with u and substitute $sec^{2}\theta$d$\theta$ with dv. So v= $\int$$sec^{2}\theta$d$\theta$=tan$\theta$
$\int$$sec^{5}\theta$d$\theta$=($sec^{3}\theta$)(tan$\theta$)-3$\int$$tan^{2}$$\theta$ $sec^{3}\theta$d$\theta$
Use the identity $tan^{2}$$\theta$ =$sec^{2}$$\theta$-1,
$\int$$sec^{5}\theta$d$\theta$ = ($sec^{3}\theta$)(tan$\theta$)-3$\int$($sec^{2}$$\theta$-1) $sec^{3}\theta$d$\theta$
And split the integral: $\int$$sec^{5}\theta$d$\theta$ =($sec^{3}\theta$)(tan$\theta$)-3$\int$$sec^{5}\theta$d$\theta$ +3 $\int$$sec^{3}\theta$d$\theta$
From step 3 we get $\int$$sec^{3}\theta$$d\theta$=$\frac{1}{2}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|)
So $\int$$sec^{5}\theta$d$\theta$ =($sec^{3}\theta$)(tan$\theta$)-3$\int$$sec^{5}\theta$d$\theta$ +$\frac{3}{2}$(sec$\theta$·tan$\theta$+ln|tan$\theta$+sec$\theta$|)
Solving this equation about $\int$$sec^{5}\theta$d$\theta$, we have
$\int$$sec^{5}\theta$d$\theta$=$\frac{1}{4}$$sec^{3}\theta$·tan$\theta$+$\frac{3}{8}$ln|tan$\theta$+sec$\theta$|+$\frac{3}{8}$tan$\theta$·sec$\theta$
Step 5: Plugging the results of $\int$$sec^{3}\theta$d$\theta$ and $\int$$sec^{5}\theta$d$\theta$ we got from step 3 and step 4 into A=$\frac{\pi}{4}$($\int$$sec^{5}\theta$d$\theta$-$\int$$sec^{3}\theta$d$\theta$) from step 2, we get:
A=$\pi$($\frac{1}{16}$$sec^{3}\theta$·tan$\theta$ -$\frac{1}{32}$ln|tan$\theta$+sec$\theta$| -$\frac{1}{32}$tan$\theta$·sec$\theta$)
Convert $\theta$ back to x using the trigonometry relationship in the image below:
A=$\pi$[$\frac{1}{16}$$(1+(2x)^2)^\frac{3}{2}$·(2x)-$\frac{1}{32}$ln|2x+$\sqrt{1+(2x)^2}$|-$\frac{1}{32}$(2x)$\sqrt{1+(2x)^2}$]
Finally, plug $x_{i}$=0 & $x_{f}$=1 into this expression about x, and get the final answer,
[$\frac{9\sqrt5}{16}$-$\frac{1}{32}$ln(2+$\sqrt5$)]$\pi$.
