The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part.

Question

Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.

This question is from Basic Mathematics. Please explain how I can solve it according to class 11th student.

My work is below.

$\implies y = \frac{x^2}2 \\ \implies 2y = x^2 \\ \implies -x^2 +2y = 0 \ \ \ \ (1)$

and

$x^2 + y^2 = 8 \ \ \ \ (2) $

Adding (1) and (2)

$y^2 + 2y = 8 \\ (y-2)(y+4) =0 \\ \therefore y = 2 or -4$

But when I plot it in the plain with help of https://www.desmos.com/ value of y can only be 2 for the point of intersections.

I'm unable to understand how I can eliminate -4 from the results.

Answer 1

Find the intersection of the curves $y=\frac{x^2}{2}$ with $x^2+y^2=8$

From the parabola $y=\frac{x^2}{2}$ we have $2y=x^2$. Pluging this result $x^2=2y$ in the equation function of the circle gives $2y+y^2=8$. The equation $2y+y^2=8$ is a quadratic equation $y^2+2y-8=0$ and can be solved by many methods. One of these methods is completing the square: $y^2+2y-8=(y+1)^2-9=0$.This last equation can be brought to the form $(y+1)^2=9$. Taking the square root of both sides, we get $y_{1,2}+1 = \pm 3$ and subtracting one from both sides of the equation we get $y_{1,2} = -1\pm 3$. This gives finally two possible solutions for the original problem: $y_1 = -1+3=+2 $ and $y_2 =-1-3= -4$. A possible solution is a "maybe" solution, but not yet a solution!

Then check, if $y_1=+2$ is a solution: Plug $y_1=+2$ in the parabola equation $y=\frac{x^2}{2}$ to get $2=\frac{x^2}{2} \Leftrightarrow x^2 = 4\Rightarrow x_{1,2} = \pm 2$. Plug $y_1$ in the circle equation $x^2+y^2=8$ to get $x^2+4=8 \Leftrightarrow x^2 = 4 \Rightarrow x_{1,2} = \pm 2$ (or $x_1 = +2$ and $x_2 = -2$). This means that the points $(x_{1},y_1)=(+2,+2)$ and $(x_{2},y_1)=(+2,-2)$ lie on both curves, so these are intersection points of the parabola and the circle. Then also check, if $y_2=-4$ is a solution: Plug $y_2=-4$ in the parabola equation $y=\frac{x^2}{2}$ to get $-4=\frac{x^2}{2} \Leftrightarrow x^2 = -8\Rightarrow x = \sqrt{-8} = \sqrt{8}~i= 2\sqrt{2}~i$ with the imaginary unit number $i$. In this problem we are not dealing with the complex plane, that's why, $y_2=-4$ is no good solution.

Now if you want the smaller area $A$, then substract the $y-$coordinates ($f(x)=y_{\text{circle}}-y_{\text{parabola}}$) and integrate over this function $f(x)$ from the left to the right intersection point along the $x$-axis or from $x=-2$ to $x=+2$ like this:

$$A=\int_{-2}^{2}f(x)~dx =\int_{-2}^{2}(y_{\text{circle}}-y_{\text{parabola}})~dx=\int_{-2}^{2}(\sqrt{8-x^2}-\frac{x^2}{2})~dx = 2\pi+\frac{4}{3}.$$

Answer 2

Since we have $$y=2x^2$$ so $$y=-4$$ can not be a solution,$$-8=x^2$$

Answer 3

Because $y$ should be positive.

By your work it's: $$2\int\limits_0^2\left(2-\frac{x^2}{2}\right)dx+\frac{1}{4}\cdot\pi(2\sqrt2)^2-\frac{1}{2}(2\sqrt2)^2.$$ I got $\frac{4}{3}+2\pi.$

Answer 4

I can substitute: $(x^2/2)^2+x^2=8$. I obtain: $x^4+4x^2-32=0$. Let $t=x^2$: $t^2+4t-32=0$. In other words: $(t+8)(t-4)$. $x^2=-8$ is impossible; $x^2=4$ has two solutions: $x=2$ or $x=-2$