MHT CET · Maths · Area Under Curves
The area of the region include between the parabolas \(y^2=8 x\) and \(\mathrm{x}^2=8 \mathrm{y}\), is
- A \(\frac{128}{3}\) sq. units
- B \(\frac{64}{3}\) sq. units
- C \(\frac{32 \sqrt{8}}{3}\) sq. units
- D \(\frac{16 \sqrt{8}}{3}\) sq. units
Answer & Solution
Correct Answer
(B) \(\frac{64}{3}\) sq. units
Step-by-step Solution
Detailed explanation
Refer figure
Required area is shaded. Point of intersection of given curves are \(\mathrm{y}^2=8 \mathrm{x}\) and \(\mathrm{x}^2=8 \mathrm{y}\) i.e.
\(
\begin{aligned}
& \left(\frac{x^2}{8}\right)^2=8 x \Rightarrow x\left(x^3-512\right)=0 \\
& \therefore O \equiv(0,0) \text { and } P \equiv(8,8) \\
& A=\int_0^8(2 \sqrt{2})(\sqrt{x}) d x-\int_0^8 \frac{x^2}{8} d x
\end{aligned}
\)
\(\left.=\frac{2 \sqrt{2}}{\left(\frac{3}{2}\right)} x^{\frac{3}{2}}\right]_0^8-\frac{1}{24}\left[x^3\right]_0^8=\left(\frac{4 \sqrt{2}}{3}\right)(8 \sqrt{8})-\frac{1}{24}(512)\) \(=\frac{64}{3} \text { sq. units }\)
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