Question

Area enclosed by curve \(y = 2x^2\) and lines \(x\geq 1, y\leq 4\) is ..... sq. units

• A. \(\frac{8\sqrt{2}-10}{3}\)
• B. \(\frac{8(\sqrt{2}-1)}{3}\)
• C. \(\frac{4\sqrt{2}-5}{3}\)
• D. \(\frac{4(\sqrt{2}-1)}{3}\)

Hint:

The curve meets \(y=4\) at \(x=\sqrt2\), so integrate \(4-2x^2\) from 1 to \(\sqrt2\).

Answer 1

The Correct Option is A

Step 1: Slice horizontally
For $y$ from 2 to 4, $x$ runs from 1 to $\sqrt{y/2}$.

Step 2: Integrate
$\int_2^4\left(\sqrt{y/2}-1\right)dy=\dfrac{1}{\sqrt2}\cdot\dfrac23\left(8-2\sqrt2\right)-2=\dfrac{16-4\sqrt2}{3\sqrt2}-2=\dfrac{8\sqrt2-4}{3}-2=\dfrac{8\sqrt2-10}{3}$, option (A).

Final Answer:
Area $=\frac{8\sqrt2-10}{3}$, option (A). \[ \boxed{\dfrac{8\sqrt2-10}{3}} \]