Area of circle and parabola I shall determine the surface area of the intersection of a circle with Radius $R= 1 $ and the set of points above the parabola $f(x) = 2x^2$. Only the positive $x,y$-plane is of interest.
My approach is to parameterize it with polar coordinates as $x=R\cos(\phi)$ and $y=R\sin(\phi)$, thus  $\vec \gamma(\phi) =  \begin{pmatrix}
        R \cos(\phi) \\
        R \sin(\phi) \\
        0 \\
        \end{pmatrix}$
Now I struggle with setting bounds of the integral, something like $0 \le R \le 1 $ and $? \le \phi \le ?.$ 
I have determined the point of intersection as $x=0.62481$, thus $\phi = 51.332°$ or $\phi = 0.896$ in rad. How do I continue? 
 A: I will assume the circle is centered at the origin since you did not state.
We will have the equation of the circle: $x^2+y^2=1$
For the parabola, we have $y=2x^2$
To get the points of intersection, we just have to set y=$2x^2$ in the equation of the parabola.
$x^2+4x^4=1$
Solving the equation, we get $x=0.68,-0.68$ as you said.
You should note that $y=2x^2$ is above the x-axis and thus we can take only the upper semicircle, and thus $y=\sqrt{(1-x^2)}$
Now, to continue, all we have to do is compute the integral $\int_{0}^{0.68} (\sqrt{(1-x^2)}-2x^2\,dx$
We set $x=\sin(\theta)$ so $dx=\cos(\theta)\,d\theta$ When $x=0,\theta=0$ and $x=0.68,\theta=0.74$ so the limits of integration are set.
We now calculate the indefinite integral:
\begin{align}
& \int \sqrt{1-x^2}-2x^2\,dx=\int (\sqrt{1-\sin^2 \theta} -2\sin^2(\theta))\cos(\theta) \, d\theta \\[8pt]
= {} & \int \cos^2(\theta) \,d\theta-2\int \sin^2 \theta \cos(\theta)\,d\theta=\frac{1}{2} \int 1+\cos(2\theta)\,d\theta \\[6pt]
& {} -2\cdot\frac{1}{4}\int \cos(\theta)-\cos(3\theta)\,d\theta=\frac{1}{2}\theta + \frac{1}{4} \sin(2\theta)-\frac{1}{2}\sin(\theta)+\frac{1}{6}\sin(3\theta)
\end{align}
Now, putting the values at the limit of integration, we get:
Area=0.41
A: here is way to find the area without polar coordinates. the parabola $y = 2x^2$ and the unit circle cut at $(\pm \sqrt{b/2}, b)$ where $b = {\sqrt{17} - 1  \over 4}$ is the positive solutions of the quadratic equation $y^2 + {y \over 2} = 1$
the area is $$2\int_0^b \sqrt{y \over 2} \ dy + \arccos b - b\sqrt{1 - b^2}={2\sqrt 2 \over 3} b^{3/2} +\arccos b - b\sqrt{1 - b^2}$$ 
