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Q.
The area above the x-axis enclosed by the curves $x^2-y^2 = 0$ and $x^2 + y - 2 = 0$ is
Application of Integrals
Solution:
We first draw the given curves
The first cuxrve $x^{2} - y^{2} = 0 \Rightarrow y = \pm x$ represents a pair of straight lines with slopes 1 and -1 passing through origin. The second curve
$\Rightarrow x^{2} + y - 2 = 0 \Rightarrow x^{2} = -y + 2 \Rightarrow x^{2} = -\left(y-2\right)$
represents a parabola with vertex (0,2) axis as y-axis and concavity dawnwards (see the chapter of parabola in coordinates). Both the curves are plotted in the figure and the required area is shown by the shaded region.
The points A and C are the points of intersection of $y^2 = x^2$ with $x^2 + y - 2 = 0.$
Solving the two equations, we get $y^2 + y - 2 = 0$ [putting value of $x^2 = y^2$]
$\Rightarrow \left(y+2\right)\left(y-1\right) = 0$
giving y = -2 and 1, but y = -2 is discarded as the required area is above the x-axis.
$\therefore y = 1 \Rightarrow x = \pm1$
The points A and C are respectively (-1, 1) and (1, 1) now due to symmetry
Area of the bounded region OABCO
$= 2 ×$ Area $OBCO = 2x \int^{1}_{0} \left[\left(2-x^{2}\right)-x\right]dx$
[Since $y = 2 -x^2$ is the upper curve and $y = x$ is the lower curve]
$= 2\left[2x-\frac{x^{3}}{3}-\frac{x^{2}}{2}\right]^{1}_{0} = 2\left[2-\frac{1}{3}-\frac{1}{2}\right] = \frac{7}{3}$
