Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the part of the circle $x^2+y^2=36$, which is outside the parabola $y^2=9x$, is :

We are tasked with finding the area of the region inside the circle $x^2+y^2=36$ and outside the parabola $y^2=9x$. The circle has a radius of $r=6$ and is centered at the origin.

To find the points of intersection, we substitute $y^2=9x$ into the circle's equation:

Factoring the quadratic equation, we obtain:

Since the parabola $y^2=9x$ only exists for $x\ge 0$, we discard the solution $x=-12$. Thus, the curves intersect at $x=3$.

The area of the region inside the circle but outside the parabola is found by subtracting the area enclosed by the parabola from the total area of the circle. The total area of the circle is:

Because the region is symmetric about the $x$-axis, we calculate the area in the first quadrant and multiply the result by $2$.

In the first quadrant, the region enclosed by the parabola and the circle is bounded by two different functions. From $x=0$ to $x=3$, the upper boundary is the parabola $y=3\sqrt{x}$. From $x=3$ to $x=6$, the upper boundary is the circle $y=\sqrt{36-x^2}$.

We calculate the area under these curves in the first quadrant as follows:

Evaluating the first integral:

Evaluating the second integral using the standard form $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}(\frac{x}{a})$:

Simplifying this expression yields:

The total area inside the parabola is twice the sum of these two integrals:

Finally, we subtract this from the total area of the circle to find the area of the region outside the parabola: