Find the area of the circle $x^2+y^2=16$ which is exterior to the parabola $y^2=6x$ using definite integral.
My attempt-
I have solved two to find intersection point $(2,\sqrt{12})$.
But when its come to exterior area. I have no clue. Please help me.
A non degenerated parabola divides the plane in two regions. Only one of them is convex. The convex region is often called the "interior" of the parabola.
Since the area has the $X$ axis as a symmetry axis, you can find the area above and multiply it by $2$.
This part can be divided in two: a quarter of circle (at left) whose area can be found with the known formula $\pi r^2$, or if you are not allowed to use this kind of formulas, with the integral $$\int_{-4}^0f(x)dx$$ and the "curved triangle" at right, that is between the circle and the parabola, which can be found with the integral $$\int_0^2(f(x)-g(x))dx$$
where $f$ is the function that describes the circle and $g$ is that of the parabola.
Note that you must solve the respective equations for $y$ in order to find appropiate expressions for $f$ and $g$.
Area bounded by two regions,
= $\int_0^2 \sqrt{6x} \,dx + \int_2^4 \sqrt{16-x^2}\,dx$
Then multiply this area with 2 as it is symmetric.
Area of circle = $πr^2$
= $π(4)^2$
= 16π
Required exterior area = Area of circle - Area bounded by two regions.
The exterior of a conic is the set of points from where you can draw tangents to it (including asymptotes for the hyperbola).
The drawing should make it clear what region you're concerned with. Of course you can just compute the area delimited by the arc $OA$, the arc $AB$ and the segment $OB$ in the drawing below and then subtract twice it from the are of the circle.
