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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.

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    Did you try drawing the parabola and circle? – 2017-01-05
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    @AdamHughes The axis of the parabola is horizontal hence "below the parabola" does not apply. – 2017-01-08
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    With anything convex I would assume they mean the area of the region on the "convex side" of it, but if you have doubts I would simply ask the person who posed the problem to you. – 2017-01-08

3 Answers 3

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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.

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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.

enter image description here

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 $g$.

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    but what is the answer and how to find? – 2017-01-05
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    Can you tell me how to find area bounded by parabola and circle using integral then I find remaining area by subtracting it from the area of a circle. – 2017-01-05
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The exterior of a conic is the set of points from where you can draw tangents to it (including asymptotes for the hyperbola).

enter image description here

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.

enter image description here

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    But how to integrate this? – 2017-01-05
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    @JohnSr $\displaystyle\int_0^2\sqrt{6x}\,dx+\int_2^4\sqrt{16-x^2}\,dx$ – 2017-01-05