Question

Area bounded between the curves \[ x = -2y^2 \] \[ x = 1 - 4y^2 \] is

• A. \(\frac{\sqrt{2}}{3}\)
• B. \(\frac{2\sqrt{2}}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3\sqrt{3}}{3}\)

Hint:

If curves are expressed as \(x=f(y)\), integrate with respect to \(y\) using \(A=\int(x_{right}-x_{left})dy\).

Answer 1

The Correct Option is B

Concept: When curves are given in the form \(x=f(y)\), area between them is: \[ A = \int (x_2 - x_1)\,dy \] where \(x_2\) is the right curve and \(x_1\) is the left curve.
Step 1: Find intersection points \[ -2y^2 = 1 - 4y^2 \] \[ 2y^2 = 1 \] \[ y^2 = \frac{1}{2} \] \[ y = \pm \frac{1}{\sqrt{2}} \]
Step 2: Determine right and left curves Right curve: \[ x = 1 - 4y^2 \] Left curve: \[ x = -2y^2 \]
Step 3: Set up the integral \[ A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} [(1-4y^2)-(-2y^2)]\,dy \] \[ A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (1-2y^2)\,dy \]
Step 4: Use symmetry Since the function is even, \[ A = 2\int_{0}^{\frac{1}{\sqrt{2}}} (1-2y^2)\,dy \]
Step 5: Evaluate \[ A = 2\left[y-\frac{2y^3}{3}\right]_{0}^{\frac{1}{\sqrt{2}}} \] \[ A = 2\left(\frac{1}{\sqrt{2}}-\frac{2}{3}\frac{1}{2\sqrt{2}}\right) \] \[ A = 2\left(\frac{1}{\sqrt{2}}-\frac{1}{3\sqrt{2}}\right) \] \[ A = 2\left(\frac{2}{3\sqrt{2}}\right) \] \[ A = \frac{4}{3\sqrt{2}} \] \[ A = \frac{2\sqrt{2}}{3} \]