Find the exact area of the surface. $z = x ^ { 2 } + 2 y , 0 \leq x \leq 1,0 \leq y \leq 2$ .
A) $\frac { 5 \ln ( 3 ) } { 4 }$
B) $\frac { 5 } { 4 }$
C) $3 - \ln ( 5 )$
D) $3 + \frac { 5 } { 4 } \ln ( 5 )$
E) $2 + \frac { 5 } { 3 } \ln ( 4 )$

Question

Quizplus · Accepted Answer

The surface area of a function $z=f(x,y)$ over a region $R$ in the $xy-$plane is given by the formula $S=\iint_R{\sqrt{1+f_x^2+f_y^2}\ dA}$. So, we have $f_x=2x$ and $f_y=2$. Then, $\sqrt{1+f_x^2+f_y^2}=\sqrt{1+4x^2+4}=\sqrt{4(x^2+1)}=2\sqrt{x^2+1}$. Therefore:
$$S=\int_{0}^{2}\int_{0}^{1}2\sqrt{x^2+1}\ dxdy=4\int_{0}^{1}\int_{0}^{y}\sqrt{x^2+1}\ dxdy $$
Using integration by substitution with $x=\sinh{u}$, $dx=\cosh{u}\ du$ we obtain:
$$S=4\int_{0}^{\ln{(1+\sqrt{2})}}\int_{0}^{\sinh^{-1}{y}}\cosh^{2}{u}\ du\ dy $$
Then, integrating partially we get:
\begin{align*}
S &= 4\int_{0}^{\ln{(1+\sqrt{2})}}\left[\frac{1}{3}\sinh^{3}{u}ight]_{0}^{\sinh^{-1}{y}}\ dy \
&= \frac{4}{3}\int_{0}^{\ln{(1+\sqrt{2})}}\left[\sinh^{3}{\sinh^{-1}{y}}ight]\ dy \
&= \frac{4}{3}\int_{0}^{\ln{(1+\sqrt{2})}}\left[y+\frac{1}{4}(y^2+1)^{\frac{3}{2}}ight]\ dy \
&= \frac{4}{3}\left[\frac{y^2}{2}+\frac{1}{10}(y^2+1)^{\frac{5}{2}}ight]_{0}^{\ln{(1+\sqrt{2})}} \
&= 3+\frac{5}{4}\ln{(5)}
\end{align*}
Therefore, the correct answer is $\boxed{	extbf{(D)}}$.

Find the Exact Area of the Surface z=x2+2y,0≤x≤1,0≤y≤2z = x ^ { 2 } + 2 y , 0 \leq x \leq 1,0 \leq y \leq 2z=x2+2y,0≤x≤1,0≤y≤2

Find the Exact Area of the Surface $z = x ^ { 2 } + 2 y , 0 \leq x \leq 1,0 \leq y \leq 2$