Question

Draw a rough sketch of the region {(x, y) : y² ≤ 3x, 3x² + 3y² ≤ 16} and find the area enclosed by the region using method of integration.

Answer 1

The  given region is the intersection of \[y^2 \leq 3x\text{ and }3 x^2 + 3 y^2 \leq 16\]

Clearly ,y² = 3x is a parabola with vertex at (0, 0) axis is along the x-axis opening in the positive direction.
Also 3x² + 3y² = 16 is a circle with centre at origin and has a radius \[\sqrt{\frac{16}{3}}\]

Corresponding equations of the given inequations are

\[y^2 = 3x . . . . . \left( 1 \right)\]

and
\[3 x^2 + 3 y^2 = 16 . . . . . \left( 2 \right)\]

 Substituting the value of y² from (1) into (2)

\[3 x^2 + 9x = 16\]

\[\Rightarrow 3 x^2 + 9x - 16 = 0\]

\[\Rightarrow x = \frac{- 9 \pm \sqrt{81 + 192}}{6}\]

\[\Rightarrow x = \frac{- 9 \pm \sqrt{273}}{6}\]

By figure we see that the value of x will be non-negative.

\[\therefore x = \frac{- 9 + \sqrt{273}}{6}\]

Now assume that x-coordinate of the intersecting point,

\[a = \frac{- 9 + \sqrt{273}}{6}\]

The  Required area A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length
\[= \left| y_1 \right|\]width = dx
Area of   OACO \[= \int_0^a \left| y_1 \right| d x\]
\[= \int_0^a y_1 d x\]
\[= \int_0^a \sqrt{3x} d x ...........\left( \because {y^2}_1 = \sqrt{3x} \Rightarrow y_1 = \sqrt{3x} \right)\]
\[= \left[ \frac{2\sqrt{3} x^\frac{3}{2}}{3} \right]^a_0\]
\[= \frac{2\sqrt{3} a^\frac{3}{2}}{3}\]
Therefore, Area of  OACO \[= \frac{2\sqrt{3} a^\frac{3}{2}}{3}\]

Similarly approximating the are of CABC the length \[= \left| y_2 \right|\] and the width = dx
Area of  CABC \[= \int_a^\frac{4}{\sqrt{3}} \left| y_2 \right| d x\]

\[= \int_a^\frac{4}{\sqrt{3}} y_2 dx\]

\[= \int_a^\frac{4}{\sqrt{3}} \sqrt{\frac{16}{3} - x^2} d x .............\left( \because 3 x^2 + 3 {y_2}^2 = 16 \Rightarrow y_2 = \sqrt{\frac{16}{3} - x^2} \right)\]

\[= \left[ \frac{4}{2\sqrt{3}}\sqrt{\frac{16}{3} - \frac{16}{3}} + \frac{8}{3} \sin^{- 1} \left( \frac{4\sqrt{3}}{4\sqrt{3}} \right) - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} - \frac{8}{3} \sin^{- 1} \left( \frac{a\sqrt{3}}{4} \right) \right]\]

\[= - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} + \frac{8}{3} \sin^{- 1} \left( 1 \right) - \frac{8}{3} \sin^{- 1} \left( \frac{a\sqrt{3}}{4} \right)\]
 Area of  CABC\[= - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} + \frac{4\pi}{3} - \frac{8}{3} \sin^{- 1} \left( \frac{a\sqrt{3}}{4} \right)\] 

Thus the required area A = 2(Area of OACO + Area of CABC) 

\[= 2\left[ \frac{2\sqrt{3} a^\frac{3}{2}}{3} - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} + \frac{4\pi}{3} - \frac{8}{3} \sin^{- 1} \left( \frac{a\sqrt{3}}{4} \right) \right]\]

\[= \frac{4 a^\frac{3}{2}}{\sqrt{3}} - a\sqrt{\frac{16}{3} - a^2} + \frac{8\pi}{3} - \frac{16}{3} \sin^{- 1} \left( \frac{a\sqrt{3}}{4} \right)\]

Hence, the reguired area is,]
\[\frac{4 a^\frac{3}{2}}{\sqrt{3}} - a\sqrt{\frac{16}{3} - a^2} + \frac{8\pi}{3} - \frac{16}{3} \sin^{- 1} \left( \frac{a\sqrt{3}}{4} \right)\text{ square units , where a }= \frac{- 9 + \sqrt{273}}{6}\]