Question

Find the area described by the region $5{\mathrm{x}}^2\le \mathrm{y}\le 2{\mathrm{x}}^2+9$

• A. $4\sqrt{3}$
• B. $6\sqrt{3}$
• C. $12\sqrt{3}$
• D. $8\sqrt{3}$

Answer 1

The correct option is C

$12\sqrt{3}$

Explanation for the correct option:

Step 1: Finding the coordinates

Given: $5{\mathrm{x}}^2\le \mathrm{y}\le 2{\mathrm{x}}^2+9$

So, the two equations of the curves are as follows:

$$\begin{gathered} \mathrm{y}=5{\mathrm{x}}^2 \\ \mathrm{y}=2{\mathrm{x}}^2+9 \end{gathered}$$

Equate the two curve equations to get the coordinates.

$$\begin{gathered} 5{\mathrm{x}}^2=2{\mathrm{x}}^2+9 \\ \Rightarrow 3{\mathrm{x}}^2=9 \\ \Rightarrow \mathrm{x}=\sqrt{3} \end{gathered}$$

Substituting the value of $x$ in $\mathrm{y}=5{\mathrm{x}}^2$

$$\begin{gathered} y=5{\left(\sqrt{3}\right)}^2 \\ \mathrm{y}=15 \end{gathered}$$

So, the coordinate is $\left(\sqrt{3},15\right)$.

Step 2: Calculating the area

Estimate the area of the closed figure under specified limits.

$\begin{array}{rcl}A& =& 2{\int }_0^{\sqrt{3}}\left(2{\mathrm{x}}^2+9-5{\mathrm{x}}^2\right)d\mathrm{x}\\ & =& 2{\left(9{\mathrm{x}}^2-{\mathrm{x}}^3\right)}_0^{\sqrt{3}}\\ & =& 2\left[9\sqrt{3}-3\sqrt{3}\right]\\ & =& 12\sqrt{3}\mathrm{sq}\mathrm{units}\end{array}$

Hence, option (C) is the correct answer.