Draw a rough sketch of the region {(x,y):$y^2\leq 6ax$ and $x^2+y^2\leq 16a^2$}.Also find the area of the region sketched using method of integration.

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• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\Large\frac{x}{a}\big)$$+c Step 1: Given:{(x,y):y^2\leq 6ax and x^2+y^2\leq 16a^2} Consider x^2+y^2\leq 16a^2 or x^2+y^2=16a^2 Clearly this curve represents a circle with centre (0,0) and radius 4a Consider y^2\leq 6ax or y^2=6ax Clearly this represents a parabola whose vertex is (0,0) and open rightwards. The region whose area is required is the shaded portion shown in the fig. Step 2: Let us find the point of intersection between the two curves. y^2=16a^2-x^2 and y^2=16ax 16a^2-x^2=6ax\Rightarrow y^2=6ax 16a^2-x^2=6ax\Rightarrow x^2+6ax-16a^2=0 On factorizing we get, (x+8a)(x-2a)=0 \Rightarrow x=-8a and 2a Since x=-8a\Rightarrow y^2=6a(-8a)= -ve. This is inadmirrable,since the value of y are imaginary. Hence x=(2a,y) and (2a,-y) are the points of intersection. Step 3: Now the area of the required region A=2\int_0^{2a}(y_2+y_1)dx Here y_2=\sqrt{6ax}. and y_1=\sqrt{16a^2-x^2} Since the area is enclosed in the I and IV quadrants, Area=2\times (Area of OAB) \qquad=2\begin{bmatrix}\int_0^{2a}\sqrt{6ax}+\int_{2a}^{4a}\sqrt{16a^2-x^2}\end{bmatrix}dx. \qquad=2\int_0^{2a}\sqrt{6ax}dx+2\int_{2a}^{4a}\sqrt{16a^2-x^2}dx. \int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\Large\frac{x}{a}\big) Step 4: On integrating we get, A=2.\sqrt{6a}\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}+2\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{16a^2-x^2}+\large\frac{16a^2}{2}\normalsize \sin^{-1}\big(\large\frac{x}{4a}\big)\end{bmatrix}_{2a}^{4a} On applying limits we get, A=2\times \sqrt 2\sqrt 3 a\times\large\frac{2}{3}[\normalsize (2a)^{\Large\frac{3}{2}}-0]+2\begin{bmatrix}\large\frac{4a}{2}\normalsize\sqrt{16a^2-16a^2}+8a^2 \sin^{-1}\big(\large\frac{4a}{4a}\big)\end{bmatrix}-2\begin{bmatrix}\large\frac{2a}{2}\normalsize\sqrt{16a^2-4a^2}+\large\frac{16a^2}{2}\normalsize \sin^{-1}\big(\large\frac{x}{4a}\big)\end{bmatrix} \;\;=\large\frac{4\sqrt 2\sqrt 3 a}{3}[\normalsize 2\sqrt 2\times a\sqrt a]+[16a^2\sin^{-1}(1)]-2a(2\sqrt 3a)-16a^2\sin^{-1}\big(\large\frac{1}{2}\big) \;\;=\large\frac{16\sqrt{3}a^2}{3}$$+16a^2\large\frac{\pi}{2}$$-4a^2\sqrt 3-16a^2\large\frac{\pi}{6} \;\;=\large\frac{16\sqrt 3 a^2-12a^2\sqrt 3}{3}+$$8a^2(\pi-\large\frac{\pi}{3})$
$\;\;=\large\frac{4\sqrt 3a^2}{3}+$$8a^2\big(\large\frac{2\pi}{3}\big)=\large\frac{4}{3}$$a^2(\sqrt 3+4\pi)$
Hence the required area is $\large\frac{4}{3}$$a^2(\sqrt 3+4\pi)$
edited May 7, 2013