Question

Sketch the region common to the circle ${{x}^{2}}+{{y}^{2}}=16$ and the parabola ${{x}^{2}}=6y$. Also, find the area of the region, using integration.

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).
Given expressions in the question are written as:
$\Rightarrow {{x}^{2}}+{{y}^{2}}=16$ - (1)
$\Rightarrow {{x}^{2}}=6y$ - (2)
By basic knowledge of co – ordinary geometry we can say curve 2 is an upward parabola with vertex (0, 0). We know ${{\left( x-g \right)}^{2}}+{{\left( y-h \right)}^{2}}={{a}^{2}}$ is circle with radius a, center at (g, h). So, ${{x}^{2}}+{{y}^{2}}=16$ is a circle with center (0, 0), radius 4.
So, we can sketch them as,
Curves: - ${{x}^{2}}+{{y}^{2}}=16$

${{x}^{2}}=6y$

Combining: -

Common area shading gives.

Substituting equation (2) in equation (1), we get:
$\Rightarrow 6y+{{y}^{2}}=16$
We know roots of equation $a{{x}^{2}}+bx+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Here we have a = 1, b = 6, c = -16. By substituting these, we get:
\Rightarrow $$$$y=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\left( 1 \right)\left( -16 \right)}}{2}
By simplifying the root term, we get u values as:
$\Rightarrow y=\dfrac{-6\pm \sqrt{36+64}}{2}=\dfrac{-6\pm 10}{2}$
By simplifying we get y = 2, -8.
By equation (2) we can say y must be positive because the square of number cannot be negative. So, y = 2.
Substitute this into equation (2) we can write x as:
$\Rightarrow {{x}^{2}}=6\times 2$
By applying square root on both sides we get it as:
$\Rightarrow x=2\sqrt{3},-2$
So, marking intersection points as, $\left( -2\sqrt{3},2 \right)$ & $\left( 2\sqrt{3},2 \right)$ we can draw area as:

Area OACB is the required area which is symmetrical on y – axis.
So, area OACB = $2\times$ (area OAC with y - axis)
Area OAC with y – axis can be written as sum of area of parabola and circle from O to A and A to C respectively with y – axis.
Area OACB = $2\times$ (area of parabola from O to A + area of circle from A to C)
Area of f (y) as y varies from ${{y}_{1}}$ to ${{y}_{2}}$ is given by:
Area = $\int\limits_{{{y}_{1}}}^{{{y}_{2}}}{f\left( y \right)dy}$
Area OACB = $2\times$ $\int\limits_{O}^{A}{\left( parabola \right)dy}+\int\limits_{A}^{C}{\left( circle \right)dy}$
We know parabola: - ${{x}^{2}}=6y$. By applying square root, we get:
$\Rightarrow x=\sqrt{6y}=f\left( y \right)$
We know the circle as: ${{x}^{2}}+{{y}^{2}}=16$.
By subtracting ${{y}^{2}}$ on both sides and applying square root, we get:
$\Rightarrow x=\sqrt{16-{{y}^{2}}}=f\left( y \right)$
By diagram we can say, O = (0, 0), A = $\left( 2\sqrt{3},2 \right)$, C = (0, 4).
By substituting these all into area, we get area as:
Area = $2\times \left[ \int\limits_{0}^{2}{\sqrt{6y}dy}+\int\limits_{2}^{5}{\sqrt{16-{{y}^{2}}}dy} \right]$
We took on the y – axis to avoid terms like $2\sqrt{3}$.
By integration basic properties we know the formula:

& \Rightarrow \int{\sqrt{kx}dx}=\sqrt{k}\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C \\\ & \Rightarrow \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 \\\ \end{aligned}$$ By using these are our area expression can be written as: Area of OACB = $$2\left[ \left[ 2\dfrac{\sqrt{6}}{3}{{y}^{\dfrac{3}{2}}} \right]_{0}^{2}+\left[ \dfrac{y}{2}\sqrt{16-{{y}^{2}}}+\dfrac{16}{2}{{\sin }^{-1}}\dfrac{y}{4} \right]_{2}^{4} \right]$$ By substituting limits we get area expression to be: - Area of OACB = $$2\times \left[ \left[ \dfrac{2\sqrt{6}.2\sqrt{2}}{3} \right]-0+\left[ 2\sqrt{16-16}+\dfrac{16}{2}{{\sin }^{-1}}1-\dfrac{2}{2}\sqrt{12}-\dfrac{16}{2}{{\sin }^{-1}}\dfrac{1}{2} \right] \right]$$ $${{\sin }^{-1}}1=\dfrac{\pi }{2},{{\sin }^{-1}}\dfrac{1}{2}=\dfrac{\pi }{6}$$. By substituting these areas can be written as: Area of OACB = $$2\times \left[ \dfrac{8\sqrt{3}}{3}+8.\dfrac{\pi }{2}-2\sqrt{3}-8\dfrac{\pi }{6} \right]$$ By simplifying by taking L. C. M, we get: Area of OACB = $$2\times \left[ \dfrac{16\sqrt{3}+24\pi -12\sqrt{3}-8\pi }{6} \right]$$ By simplifying the above term, we get area as: Area of OACB = $$\dfrac{16\pi +4\sqrt{3}}{3}$$ Therefore the area bounded by 2 given curves is $$\left( \dfrac{16\pi +4\sqrt{3}}{3} \right)$$ square units. Note: Don’t forget to take the negative root for x because we have 2 roots. Alternate method is to drop the point $$\left( 2\sqrt{3},2 \right)$$ on x – x-axis as $$\left( 2\sqrt{3},0 \right)$$. Now, integrate the circle up to this point and subtract the area of parabola from O to this point. You will get the same result but calculation is hard as it involves $$\sqrt{3}$$ terms. So, do carefully to arrive at the correct result.