We have to find the area of the region $R$ between $r=\theta $ and $r=e^{\theta/2}$, and between $\theta =0$ and $\theta =\pi /2$. Since the Jacobian of the transformation from Cartesian to polar coordinates is $r$, the area $A$ is given by the double integral (see Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$):
$$
\begin{eqnarray*}
A &=&\iint_{R}dA=\int_{0}^{\pi /2}\left( \int_{\theta }^{e^{\theta
/2}}r\;dr\right)\ d\theta \\
&=&\int_{0}^{\pi /2}\big( \frac{e^{\theta }}{2}-\frac{\theta ^{2}}{2}
\big)\ d\theta \\
&=&\frac{1}{2}\int_{0}^{\pi /2}e^{\theta }d\theta -\frac{1}{2}\int_{0}^{\pi
/2}\theta ^{2}\ d\theta \\
&=&\big( \frac{e^{\pi /2}}{2}-\frac{1}{2}\big) -\frac{\pi ^{3}}{48}.
\end{eqnarray*}
$$
$y=2x^2$ has derivative $y'=4x$. So the derivative will be positive for all $x>0$. This means the tangent line will intersect the x-axis at some point $x_a<x$ for a given x.
This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $x<x_a$ which is made up of only area under the parabola, and a region $x>x_a$ where area is from the area between the parabola and the tangent to $(3,18)$.
The equation of a tangent line at $(x_0,y_0)$ is $$\frac{y-f(x_0)}{x-x_0}=f'(x_0)$$
Or : $y=f'(x_0)(x-x_0)+f(x_0)$
The x intercept happens where $y=0$.
Requiring $y=0$ implies an x intercept of $x_c=\frac{-f(x_0)}{f'(x_0)}+x_0$
So from the above arguments with $x_0=3$:
$$A=\int_0^{x_c}2x^2dx+\int_{x_c}^32x^2-(12x-18)dx$$
But this can be simplified. The area under the tangent line is a triangle. So the the parabola can be integrated ignoring the tangent line, and then subtracting the area of the triangle.
From the above, we know $(x_0-x_c)=\frac{f(x_0)}{f'(x_0)}$, the base of the triangle. The height is just $f(x_0)$.
So the area of the triangle is $A_t=\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$
So:
$$A=\int_0^{x_0}f(x)dx-\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$$
and solve for $x_0=3$.
In this form, an expression can be found for $x_0$ which extremizes the area.
Best Answer
Hints:
$x_B$ can be found by solving $3x-x^2=0$. Now you can work out the area of the right triangle colored red in the figure.
Now the area of the shaded region can be represented as
$$A=\int_{x_A}^{x_B} 3x-x^2\,dx-\text{area of the triangle}$$