Question

Using integration, find the area of the region \left\\{ \left( x,y \right):{{y}^{2}}\le 4x,4{{x}^{2}}+4{{y}^{2}}\le 9 \right\\}.

Answer 1

Hint: First, from the given expression / conditions, find 2 curves which we need to find the area. Next by using elimination method find the points of intersection of the curves. By looking at their equations, use normal co – ordinate geometry knowledge to find their shapes. Draw the curves together on a graph. By using their diagrams mark the area which is common in both of them. Now, integrate the curves to find the area and subtract the extra area to get the area which is in common to both curves.

Complete step-by-step answer:
Elimination method:
First write 2 curve equations which you need to solve to find intersection points. Then try to convert one variable in terms of another variable by using any one of the equations. Now substitute this variable in terms of others into the remaining equation. Now the remaining equation turns into an equation of one variable. By normal geometry find the variable. Using previous relations, find the other variable. Thus, you get the intersection point(s).
So, intersection points are given by $\left( \dfrac{1}{2},\sqrt{2} \right),\left( \dfrac{1}{2},-\sqrt{2} \right)$.
Now, we will draw curves and mark areas.
Curves: -
${{y}^{2}}=4ax$, parabola at origin on positive x – axis

${{x}^{2}}+{{y}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}}$, we got this by dividing 4 on both sides of (2).

Areas: - ${{y}^{2}}\le 4x$, $\le$ $\Rightarrow$ bounded inside.

${{x}^{2}}+{{y}^{2}}\le {{\left( \dfrac{3}{2} \right)}^{2}}$

Combining both the areas we can represent the bounded as,

By diagram we can say OACB is symmetric about the x – axis.
So, if we find the area from O to A by parabola and then A to C by circle we get 2 areas.
Now if you add 2 areas you will get an area of OAC with x – axis.
As it is a symmetric area of OBC = area of OAC.
Area of OACB = area of OBC + area of OAC
From above we can say that the required area,
Area of OACB = 2 (area of OAC)
The expressions / inequalities given in the question are:
${{y}^{2}}\le 4x,4{{x}^{2}}+4{{y}^{2}}\le 9$
By general co – ordinary geometry we can say curve – 1 is a parabola and curve – 2 is a circle.
For finding intersection points we write them as:
$\Rightarrow {{y}^{2}}=4x$ - (1)
$\Rightarrow 4{{x}^{2}}+4{{y}^{2}}=9$ - (2)
By applying square root on both sides to equation (1), we get:
$\Rightarrow y=2\sqrt{x}$ - (3)
By substituting equation (3) into equation (2), we get:
$\Rightarrow 4{{x}^{2}}+4{{\left( 2\sqrt{x} \right)}^{2}}=9$
By subtracting 9 on both sides of equation, we get:
$\Rightarrow 4{{x}^{2}}+16x-9=0$
We know, roots of equation $a{{x}^{2}}+bx+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
By using above, we know: a = 4, b = 16, c = -9. By substituting twe get:
$\Rightarrow x=\dfrac{-16\pm \sqrt{{{16}^{2}}-4\left( 4 \right)\left( -9 \right)}}{8}=\dfrac{-16\pm 20}{8}$
By simplifying the x value (s) we get them as, $x=\dfrac{1}{2},\dfrac{-9}{2}$.
By substituting these x into the equation (3), $x=\dfrac{-9}{2}$ is not possible because the square root of negative number is not possible. So, substitute $x=\dfrac{1}{2}$ in (3), $y=2\sqrt{\dfrac{1}{2}}$.
By simplifying we get y values as: $y=\pm \sqrt{2}$.
From before explanation we can say area of OAC as:
Area of OAC = (Area of parabola from O to A) + (A to C by circle).
Now we will find them both separately in below 2 cases.
Case 1: Area bounded by parabola from O to A with x – axis parabola is given by ${{y}^{2}}=4ax$. Let this area be P. By applying square root on both sides of equation, we get:
\Rightarrow $$$$y=2\sqrt{x}
Here from points O to A x varies as 0 to $\dfrac{1}{2}$.
So, area of curve f (x) while x varies from ${{x}_{1}}$ to ${{x}_{2}}$ is,
$\Rightarrow A=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{f\left( x \right)dx}$
Here we have $f\left( x \right)=2\sqrt{x}$, ${{x}_{1}}$ = 0, ${{x}_{2}}$ = $\dfrac{1}{2}$, from above we get:
$\Rightarrow$ Area = P = $\int\limits_{0}^{\dfrac{1}{2}}{2\sqrt{x}dx}=\int\limits_{0}^{\dfrac{1}{2}}{2{{\left( x \right)}^{\dfrac{1}{2}}}dx}$
By basic integration we can say that: $\int{k{{x}^{n}}dx}=k\left( \dfrac{{{x}^{n+1}}}{n+1}+C \right)$.
By above formula we can write P as follows:
$\Rightarrow P=2\left[ \dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]_{0}^{\dfrac{1}{2}}$
By substituting 0, $\dfrac{1}{2}$ and subtracting we get P as follows:
$\Rightarrow P=2\left( \dfrac{2}{3}{{\left( \dfrac{1}{2} \right)}^{\dfrac{3}{2}}}-0 \right)$
By simplifying, we get value of P as:
$\Rightarrow P=2\times \dfrac{2}{3}\times \dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{3}$
Case: 2 Area bounded by circle from A to C. Let it be Q.
Circle equation given as: ${{x}^{2}}+{{y}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}}$.
By subtracting ${{x}^{2}}$ and applying square root on both sides, we get:
$\Rightarrow y=\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}-{{x}^{2}}}$
From area of curve formula used in previous case, we get,
$\Rightarrow Q=\int\limits_{{{x}_{1}}}^{{{x}_{1}}}{f\left( x \right)dx}$
Here, $f\left( x \right)=\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}-{{x}^{2}}},{{x}_{1}}=\dfrac{1}{2},{{x}_{2}}=\dfrac{3}{2}$.
$\Rightarrow Q=\int\limits_{\dfrac{1}{2}}^{\dfrac{3}{2}}{\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}-{{x}^{2}}}dx}$
By basic integration: $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C$.
By this we get: $Q=\left[ \dfrac{x}{2}\sqrt{\dfrac{9}{4}-{{x}^{2}}}+\dfrac{9}{2\times 4}{{\sin }^{-1}}\dfrac{x}{\dfrac{3}{2}} \right]_{\dfrac{1}{2}}^{\dfrac{3}{2}}$.
$\Rightarrow Q=\dfrac{3}{4}\sqrt{\dfrac{9}{4}-\dfrac{9}{4}}+\dfrac{9}{8}{{\sin }^{-1}}1-\dfrac{1}{4}\sqrt{\dfrac{9}{4}-\dfrac{1}{4}}+\dfrac{9}{8}{{\sin }^{-1}}\dfrac{1}{3}$
By simplifying we get Q value as:
$\Rightarrow Q=\dfrac{9\pi }{16}-\dfrac{\sqrt{2}}{4}-\dfrac{9}{8}{{\sin }^{-1}}\dfrac{1}{3}$ ($\because$ we know, ${{\sin }^{-1}}1=\dfrac{\pi }{2}$)
By formula of area of OAC described before, we get:
Area of OAC = P + Q = $\dfrac{\sqrt{2}}{3}+\dfrac{9\pi }{16}-\dfrac{\sqrt{2}}{4}-\dfrac{9}{8}{{\sin }^{-1}}\dfrac{1}{3}$.
As required area is symmetric we write it as:
Required area = $2\times$ (Area of OAC) = $2\times \left( \dfrac{\sqrt{2}}{3}+\dfrac{9\pi }{16}-\dfrac{\sqrt{2}}{4}-\dfrac{9}{8}{{\sin }^{-1}}\dfrac{1}{3} \right)$
By taking L. C. M of 2 numerical inside bracket we get,
Area = $2\times \left( \dfrac{9\pi }{16}-\dfrac{9}{8}{{\sin }^{-1}}\dfrac{1}{3}+\dfrac{\sqrt{2}}{12} \right)$.
By simplifying more we can write it as:
Area = $\dfrac{9\pi }{8}-\dfrac{9}{4}{{\sin }^{-1}}\dfrac{1}{3}+\dfrac{\sqrt{2}}{6}=\dfrac{9\pi }{8}-\dfrac{9}{4}{{\sin }^{-1}}\dfrac{1}{3}+\dfrac{1}{3\sqrt{2}}$
Therefore area bounded by ${{x}^{2}}+{{y}^{2}}\le \dfrac{9}{4},{{y}^{2}}\le 4x$ is given by,
$\Rightarrow \left( \dfrac{9\pi }{8}-\dfrac{9}{4}{{\sin }^{-1}}\dfrac{1}{3}+\dfrac{1}{3\sqrt{2}} \right)$ square units.

Note: Be careful you must not take x value negative while calculating intersection points. And do not forget to take negative not while calculating y – value because you must get 2 intersection points. Generally students confuse between $\le ,\ge$ while taking the area. Be careful at these points.