Area of circle and parabola

Question

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?

Answer 1

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 $$2\int_0^b \sqrt{y \over 2} \ dy + \arccos b - b\sqrt{1 - b^2}={2\sqrt 2 \over 3} b^{3/2} +\arccos b - b\sqrt{1 - b^2}$$

Answer 2

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