Question

using integral find the area of that part of circle x2+y2=16, which is exterior to the parabola y2=6x.

Answer 1

Answer on Question#39220 – Math – Other

using integral find the area of that part of circle $x^2 + y^2 = 16$, which is exterior to the parabola $y^2 = 6x$.

Solution:

We have to find the area of that part of circle $x^2 + y^2 = 16$, which is exterior to the parabola $y^2 = 6x$. So, we have to find the shaded region area.

Firstly, we will find the point of intersection,

$$x^2 + 6x = 16$$ $$x^2 + 6x - 16 = 0$$ $$(x + 8)(x - 2) = 0 \Rightarrow$$ $$x = -8; 2$$

So,

$$y^2 = 6x(-8) = -48 \Rightarrow y = -\sqrt{48}$$

And

$$y^2 = 6x(2) = 12 \Rightarrow y = \sqrt{12} \Rightarrow y = \pm 2\sqrt{3}$$

So, the point of intersection are

$$(2, 2\sqrt{3}), (2, -2\sqrt{3}) \text{ and } (-8, -\sqrt{48})$$

We will take only first and second point because third point has imaginary value.

Required area = area of circle – area of the part which is not shaded

So,

$$\text{Required area} = \pi(4)^2 - 2\int_0^2 \sqrt{6x} \, dx - 2\int_2^4 \sqrt{16 - x^2} \, dx$$ $$\int_0^2 \sqrt{6x} \, dx = \sqrt{6} \left( \frac{x^2}{\frac{3}{2}} \right) \frac{2}{0} = \frac{2\sqrt{6}}{3} (2)^{\frac{3}{2}} = \frac{8\sqrt{3}}{3}$$

And

$$\int_2^4 \sqrt{16 - x^2} \, dx$$ $$\text{let } x = 4\sin\theta \Rightarrow dx = 4\cos\theta \, d\theta$$ $$\begin{array}{l} \text{limits are } 2 = 4 \sin \theta \Rightarrow \theta = \frac{\pi}{6} \text{ and } 4 = 4 \sin \theta \Rightarrow \theta = \frac{\pi}{2} \\ \Rightarrow \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sqrt{16 - (4 \sin \theta)^2} \, 4 \cos \theta \, d\theta = 16 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\cos \theta)^2 \, d\theta = 16 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1 - \cos 2\theta}{2} \, d\theta = \\ 8 \left( \theta - \frac{\sin 2\theta}{2} \right) \frac{\frac{\pi}{2}}{6} = 8 \left( \frac{\pi}{2} - \frac{\sin \pi}{2} - \frac{\pi}{6} + \frac{\sin \frac{\pi}{3}}{2} \right) = 8 \left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) \end{array}$$

Now, substitute all values in equation (1), we get,

$$\begin{array}{l} \text{Required area} = \pi(4)^2 - 2 \left( \frac{8\sqrt{3}}{3} \right) - 2 \left( 8 \left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) \right) \\ = \pi \left( 16 - \frac{16}{3} \right) - \frac{16\sqrt{3}}{3} - 4\sqrt{3} = \\ = \pi \left( \frac{32}{3} \right) - \frac{28\sqrt{3}}{3} = \frac{4}{3} (8\pi - 7\sqrt{3}) \text{sq. units} \end{array}$$

Answer: $\frac{4}{3} (8\pi - 7\sqrt{3})$ sq. units.