Question

The area (in sq units) bounded by the curves $4y=x^2$and $2y=6-x^2$ is

• A. $8$
• B. $6$
• C. $4$
• D. $10$

Answer 1

The correct option is A

$8$

Explanation for Correct Answer:

Determining the area bounded by the curves.

$x^2=4y\to \left(i\right)$ is an upward parabola.

$2y=6-x^2$ or $x^2=6-2y\to \left(ii\right)$ is a downward parabola.

Now the graph of the given curves is as shown:

Then the point of intersection of the two curves are given by substituting equation $\left(i\right)$ in equation $\left(ii\right)$

$$\begin{gathered} 4y=6-2y \\ \Rightarrow 6y=6 \\ \Rightarrow y=1 \end{gathered}$$

Substitute $y=1$in equation $\left(i\right)$

$$\begin{gathered} x^2=4\times 1 \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \end{gathered}$$

Hence the intersection points are $\left(2,1\right)$ and $\left(-2,1\right)$

Let the downward parabola represent $Y_1$ hence

$$\begin{gathered} Y_1=\frac{6-x^2}{2} \\ \Rightarrow Y_1=3-\frac{x^2}{2} \end{gathered}$$

And the upward parabola is represent $Y_2$

$$\begin{gathered} 4Y_2=x^2 \\ \Rightarrow Y_2=\frac{x^2}{4} \end{gathered}$$

Now the area of the region is given by

$\begin{array}{l}A=2{\int }_0^2(Y_1-Y_2)dx\\ =2{\int }_0^2\left(3-\frac{x^2}{2}-\frac{x^2}{4}\right)dx\\ =2{\left[3x-\frac{x^3}{6}-\frac{x^3}{12}\right]}_0^2\\ =2\left[\left(3\times 2-\frac{2^3}{6}-\frac{2^3}{12}\right)-0\right]\\ =2\left[6-\frac{8}{6}-\frac{8}{12}\right]\\ =2\left[\frac{72-16-8}{12}\right]\\ =\frac{2\times 48}{12}\\ =8\end{array}$

Therefore the area of a graph is equal to $8\text{square units}$

Hence, the correct answer is option (A)