Question

Find the area lying above X-axis and included between the circle $ {x^2} + {y^2} = 8x $ and inside the parabola $ {y^2} = 4x. $

Answer 1

Hint: $ \int\limits_a^b {f(x)dx} $ represents area under the curve $ f(x) $ between the points $ x = a $ and $ x = b $ above X-axis

Complete step-by-step answer:
Since equation of circle with center $ (h,k) $ and radius $ r $ is given by equation,
 $ {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} $
We convert our equation to the above form to get the coordinates of the center of the circle and the radius of the circle.
We have
 $ {x^2} + {y^2} = 8x $ . . . (1)
Rearranging it, we get
 $ {x^2} - 8x + {y^2} = 0 $
 $ \Rightarrow {x^2} - 2 \times 4 \times x + {y^2} = 0 $
Add $ {4^2} $ to both the sides
 $ \Rightarrow {x^2} - 2 \times 4 \times x + {4^2} + {y^2} = {4^2} $
 $ \Rightarrow {\left( {x - 4} \right)^2} + {y^2} = {4^2} $ $ \left( {\because {{(a - b)}^2} = {a^2} + 2ab + {b^2}} \right) $ . . . (2)
So, circle has centers $ \left( {4,0} \right) $ and radius $ = 4 $
Equation of parabola with vertex at the origin is given by
 $ {y^2} = 4ax $
We have the equation of parabola as
 $ {y^2} = 4x $ . . . (3)
Thus, here $ a = 1 $ and the parabola has vertex at the origin.
To find the point of intersection of parabola and circle, put the value of $ {y^2} $ from equation (3) into equation (1)
Therefore, equation (1) becomes
 $ {x^2} + 4x = 8x $
Re-arranging it, we get
 $ {x^2} + 4x - 8x = 0 $
 $ \Rightarrow {x^2} - 4x = 0 $
 $ \Rightarrow x(x - 4) = 0 $
 $ \Rightarrow x = 0 $ or $ x = 4 $
Therefore, the parabola and the circle are intersecting at points $ x = 0 $ and $ x = 4 $
Now, using equation (2), equation (3) and the point of intersections, we can draw the graph as

We need to find the area of the shaded region.
Now, we know that
 $ \int\limits_a^b {f(x)dx} $ represents area under the curve $ f(x) $ between the points $ x = a $ and $ x = b $ above X-axis
The area in the shaded region is under two curves
 $ {y^2} = 4x $ from $ x = 0 $ to $ x = 4 $
And
 $ {\left( {x - 4} \right)^2} + {y^2} = {4^2} $ from $ x = 4 $ to $ x = 8 $
Let the required area in the shaded region is given by $ A $
Then
 $ A = \int\limits_0^4 {{f_1}(x)dx} + \int\limits_4^8 {{f_2}(x)dx} $ . . . (1)
Where,
 $ {f_1}(x) = y = \sqrt {4x} $
And
 $ {f_2}(x) = y = \sqrt {{4^2} - {{\left( {x - 4} \right)}^2}} $
 $ \Rightarrow A = \int\limits_0^4 {\sqrt {4x} dx} + \int\limits_4^8 {\sqrt {{4^2} - {{\left( {x - 4} \right)}^2}} dx} $ . . . (4)
Let $ {I_1} = \int\limits_0^4 {\sqrt {4x} dx} $
 $ \Rightarrow {I_1} = 2\int\limits_0^4 {\sqrt x dx} $ $ \left( {\because \int {kf(x)dx = k\int {f(x)dx} } } \right) $
 $ \Rightarrow {I_1} = 2\int\limits_0^4 {{x^{\dfrac{1}{2}}}dx} $
 $ = 2\left[ {\dfrac{{{x^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right]_0^4 $ $ \left( {\because \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C} \right) $
 $ = 2\left[ {\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_0^4 $
Substituting the upper and lower limits and simplifying it, we get
 $ = \dfrac{4}{3}\left[ {{4^{\dfrac{3}{2}}} - 0} \right] $ $ \left( {\because \left[ {F(x)} \right]_a^b = F(b) - F(a)} \right) $
 $ = \dfrac{4}{3} \times {2^3} $
 $ \Rightarrow {I_1} = \dfrac{{32}}{3} $ . . . (5)
Let $ {I_2} = \int\limits_4^8 {\sqrt {{4^2} - {{\left( {x - 4} \right)}^2}} dx} $
We know that
 $ \int {\sqrt {{a^2} - {x^2}} dx = } \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C $
Using this formula, we can write
\[{I_2} = \left[ {\dfrac{{x - 4}}{2}\sqrt {{4^2} - {{\left( {x - 4} \right)}^2}} + \dfrac{{{4^2}}}{2}{{\sin }^{ - 1}}\left( {\dfrac{{x - 4}}{4}} \right)} \right]_4^8\]
 $ \Rightarrow {I_2} = \left[ {\dfrac{{8 - 4}}{2}\sqrt {{4^2} - {{\left( {8 - 4} \right)}^2}} + \dfrac{{{4^2}}}{2}{{\sin }^{ - 1}}\left( {\dfrac{{8 - 4}}{4}} \right) - \left( {\dfrac{{4 - 4}}{2}\sqrt {{4^2} - {{\left( {4 - 4} \right)}^2}} + \dfrac{{{4^2}}}{2}{{\sin }^{ - 1}}\left( {\dfrac{{4 - 4}}{4}} \right)} \right)} \right] $
By simplifying it, we get
 $ {I_2} = \dfrac{4}{2}\sqrt {{4^2} - {{\left( 4 \right)}^2}} + \dfrac{{{4^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{4}{4}} \right) - 0 + 0 $ $ \left( {\because {{\sin }^{ - 1}}(0) = 0} \right) $
 $ \Rightarrow {I_2} = \dfrac{{16}}{2}{\sin ^{ - 1}}(1) $
 $ \Rightarrow {I_2} = 8 \times \dfrac{\pi }{2} $ $ \left( {\because {{\sin }^{ - 1}}(1) = \dfrac{\pi }{2}} \right) $
 $ \Rightarrow {I_2} = 4\pi $ . . . (6)
By substituting the values of $ {I_1} $ and $ {I_2} $ from equation (5) and (6) into equation (4), we get
 $ A = \dfrac{{32}}{3} + 4\pi $ sq. units
Therefore, the required area is, $ \dfrac{{32}}{3} + 4\pi $ sq. units

Note: It is extremely important to note that $ \int\limits_a^b {f(x)dx} $ gives the area under the curve $ f(x) $ and above the X-axis. If the curve is under the X-axis then this integral will give a negative answer. But the area cannot be negative. Therefore, whenever the curve is under X-axis, we will write the above integral in mod. i.e. $ \left| {\int\limits_a^b {f(x)dx} } \right| $