MHT CET · Maths · Area Under Curves
\(\quad\) The area bounded by the parabola \(x^{2}=4 y\) and the lines \(y=2, \quad y=4\) and \(Y\) -axis is
- A \(\frac{4}{3}(8-2 \sqrt{2})\) sq. units
- B \(\frac{8}{3}(8-2 \sqrt{2})\) sq. units
- C \(\frac{8}{3}(8+2 \sqrt{2})\) sq. units
- D \((8-2 \sqrt{2})\) sq. units
Answer & Solution
Correct Answer
(B) \(\frac{8}{3}(8-2 \sqrt{2})\) sq. units
Step-by-step Solution
Detailed explanation
Required area is shaded.
\(\begin{aligned}
A &=2 \int_{2}^{4}(2 \sqrt{y}) d y \\
&=4 \int_{2}^{4} y^{\frac{1}{2}} d y=4\left[\frac{y^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{2}^{4}=\frac{8}{3}[4 \sqrt{4}-2 \sqrt{2}] \\
&=\frac{8}{3}(8-2 \sqrt{2})
\end{aligned}\)
\(\begin{aligned}
A &=2 \int_{2}^{4}(2 \sqrt{y}) d y \\
&=4 \int_{2}^{4} y^{\frac{1}{2}} d y=4\left[\frac{y^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{2}^{4}=\frac{8}{3}[4 \sqrt{4}-2 \sqrt{2}] \\
&=\frac{8}{3}(8-2 \sqrt{2})
\end{aligned}\)
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