Question

The area bounded by the curves \[{x^2} + {y^2} = 25\] , \[\mid 4y = \mid 4 - x2\mid \mid \] and \[x = 0\] , above x-axis is
A. $2 + \dfrac{{25}}{2}{\sin ^{ - 1}}\dfrac{4}{5}$
B. $2 + \dfrac{{25}}{4}{\sin ^{ - 1}}\dfrac{4}{5}$
C. $2 + \dfrac{{25}}{2}{\sin ^{ - 1}}\dfrac{1}{5}$
D. None of these

Answer 1

Hint: To find the area of the required region we first need to calculate the points of intersection of the two curves. We will then shade the required region and use the concept of integration to find the area of the region bounded by the two curves above the x-axis.

Formula Used: Area, \[A = \mathop \smallint \nolimits_a^b \left[ {f(x) - g(x)} \right]\:dx\]
Arithmetic Formula, ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
And $\sqrt {{a^2} - {x^2}} = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

Complete step by step Solution:
Given curves are: \[{x^2} + {y^2} = 25\] ...(1)
 \[\mid 4y = \mid 4 - {x^2}\mid \mid \] ...(2)
And \[x = 0\] ...(3)
To find the point of intersection of the curves (1) and (2), substitute the value of $y$ from equation (2) into the equation (3), we get, \[{x^2} + \dfrac{{{{\left( {4 - {x^2}} \right)}^2}}}{{16}} = 25\]
Using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ , we get, $\left( {{x^2} + 24} \right)\left( {{x^2} - 16} \right) = 0$
Thus, we get, $x = \pm 4$
 
The shaded region in the above diagram is the required region whose area is to be calculated.
If \[f\left( x \right)\] and \[g\left( x \right)\] are continuous on \[\left[ {a,{\text{ }}b} \right]\] and \[g\left( x \right){\text{ }} < {\text{ }}f\left( x \right)\] for all $x$ in \[\left[ {a,{\text{ }}b} \right]\] , then we have the following formula.
Required area, \[A = \mathop \smallint \nolimits_a^b \left[ {f(x) - g(x)} \right]\:dx\]
Here, $A = \int_0^4 {\sqrt {25 - {x^2}} dx - \int_0^2 {\left( {\dfrac{{4 - {x^2}}}{4}} \right)dx - \int_2^4 {\left( {\dfrac{{{x^2} - 4}}{4}} \right)dx} } } $
We know that, $\sqrt {{a^2} - {x^2}} = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$
Therefore, \[A = \left( {\dfrac{x}{2}\sqrt {25 - {x^2}} + \dfrac{{25}}{2}{{\sin }^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right)_0^4 - \dfrac{1}{4}\left( {4x - \dfrac{{{x^3}}}{3}} \right)_0^2 - \dfrac{1}{4}\left( {\dfrac{{{x^3}}}{3} - 4x} \right)_2^4\]
Putting the limits, we get,
 $A = \left( {\left( {2 \times 3} \right) + \dfrac{{25}}{2}{{\sin }^{ - 1}}\left( {\dfrac{4}{5}} \right) - 0} \right) - \dfrac{1}{4}\left( {8 - \dfrac{8}{3} - 0} \right) - \dfrac{1}{4}\left( {\left( {\dfrac{{64}}{3} - 16} \right) - \left( {\dfrac{8}{3} - 8} \right)} \right)$
Solving this, we get, $A = 6 + \dfrac{{25}}{2}{\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right) - \dfrac{4}{3} - \dfrac{4}{3} - \dfrac{4}{3}$
Thus, $A = 2 + \dfrac{{25}}{2}{\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right)$

Therefore, the correct option is (A).

Note: While calculating the points of intersection of the two curves, if we first find the value of $y$, then do not take the negative value of $y$ because we need to find the area of the required region above the x-axis. Also, the limits of integration should be substituted carefully. One must know the rules of integration for all such questions.