Question

The area of the region, enclosed by the circle \[{x^2} + {y^2} = 2\] which is not common to the region bounded by the parabola \[{y^2} = x\] and the straight line \[y = x\] is:
1) \[\dfrac{1}{3}\left( {15\pi - 1} \right)\]
2) \[\dfrac{1}{6}\left( {24\pi - 1} \right)\]
3) \[\dfrac{1}{6}\left( {12\pi - 1} \right)\]
4) \[\dfrac{1}{3}\left( {6\pi - 1} \right)\]

Answer 1

Hint—Construct a figure in which draw a circle of radius \[\sqrt 2 \], draw a parabola of type \[{y^2} = x\] and a straight line \[y = x\]. Find the common region bounded by the parabola and the straight line and determine the area of this region by using the integral and subtract the area of the shaded region from the area of a circle (whose formula is \[A = \pi {r^2}\]) to find the area of the region which is not common to the region bounded by the parabola and straight line.

Complete step-by-step answer:
Consider the figure below,

In the figure, we have drawn the circle, parabola and the line said in the question and have found the common region from all the three.
Now, we are given the equation of a circle which is \[{x^2} + {y^2} = 2\].
We can find the radius of a circle as the general form of the circle is \[{x^2} + {y^2} = {r^2}\]
Thus, on comparing both the forms we have,
\[r = \sqrt 2 \]
Now, we will find the area of the shaded region,
So, the function of the shaded region which is bounded above by the parabola and bounded below by the equation of a line.
Hence, subtract the region of \[{y^2} = x\] from the region of \[y = x\] with limits over integral as 0 to 1.
\[{\text{Area of the shaded region}} = \int_0^1 {\left( {\sqrt x - x} \right)dx} \]
We will evaluate the value of the integral,
\[
  {\text{Area of the shaded region = }}\left[ {\dfrac{2}{3}{x^{3/2}} - \dfrac{1}{2}{x^2}} \right]_0^1 \\
   = \left[ {\left( {\dfrac{2}{3}{{\left( 1 \right)}^{3/2}} - \dfrac{1}{2}{{\left( 1 \right)}^2}} \right) - \left( {\dfrac{2}{3}{{\left( 0 \right)}^{3/2}} -\dfrac{1}{2}{{\left( 0 \right)}^2}} \right)} \right] \\
   = \dfrac{2}{3} - \dfrac{1}{2} \\
   = \dfrac{1}{6} \\
\]
Thus, the area of the shaded region is \[\dfrac{1}{6}\].
Now, We know that the area of the circle is given by \[{\text{Area of the circle}} = \pi {r^2}\]
Here, we will substitute the value of \[r = \sqrt 2 \] and find the area,
\[
  {\text{Area of the circle}} = \pi {\left( {\sqrt 2 } \right)^2} \\
   = 2\pi \\
\]
Thus, the required area is found by subtracting the area of circle from the area of shaded region,
\[
  {\text{Required area = area of circle - area of shaded region}} \\
  {\text{ = 2}}\pi - \dfrac{1}{6} \\
   = \dfrac{{12\pi - 1}}{6} \\
\]
Thus, the area which is not common is \[\dfrac{{12\pi - 1}}{6}\].
The option (c) is the correct option.

Note: The area of the circle is \[\pi {r^2}\]. To find the function of the common area, find the region and then subtract the lower region bounded by a line from the upper region bounded by a parabola and thus, we will get the function which is to be integrated to find the value of that area. The general form of the circle is \[{x^2} + {y^2} = {r^2}\].