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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Area Under Curves

NDA 2018 · Mathematics · Area Under Curves

What is the area of the region bounded by the parabolas y^2=6(x-1)...

What is the area of the region bounded by the parabolas y2=6(x−1)y^2=6(x-1) and y2=3x?y^2=3 x ?

Options

  1. A63\frac{\sqrt{6}}{3}
  2. B263\frac{2 \sqrt{6}}{3}
  3. C463\frac{4 \sqrt{6}}{3}
  4. D563\frac{5 \sqrt{6}}{3}

Correct answer

C. 463\frac{4 \sqrt{6}}{3}

Step-by-step solution

Given parabolas are y2=6(x−1)….(1)y^2=6(x-1) \quad \ldots .(1) and y^2=3 x....(2)
y2=6(x−1)⇒6x−6=y2⇒x=y2+66=y26+1\begin{aligned} & y^2=6(x-1) \\ & \Rightarrow 6 x-6=y^2 \\ & \Rightarrow x=\frac{y^2+6}{6}=\frac{y^2}{6}+1 \end{aligned} Also, y2=3x⇒x=y23\mathrm{y}^2=3 \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{y}^2}{3}. Solving (1), (2), 3x=6(x−1)⇒3x=6x−6⇒3x=6⇒x=23 \mathrm{x}=6(\mathrm{x}-1)\Rightarrow 3 \mathrm{x}=6 \mathrm{x}-6\Rightarrow 3 \mathrm{x}=6 \Rightarrow \mathrm{x}=2. y2=6⇒y=±6y^2=6 \Rightarrow y= \pm \sqrt{6}. ∴\therefore \quad Area =∫−66(y26+1−y23)⋅dy=2∫06(1−y26)dy=2[y−y318]06=2[6−(6)318]=2×263=463\begin{aligned} & =\int_{-\sqrt{6}}^{\sqrt{6}}\left(\frac{y^2}{6}+1-\frac{y^2}{3}\right) \cdot d y=2 \int_0^{\sqrt{6}}\left(1-\frac{y^2}{6}\right) d y=2\left[y-\frac{y^3}{18}\right]_0^{\sqrt{6}} \\ & =2\left[\sqrt{6}-\frac{(\sqrt{6})^3}{18}\right]=2 \times \frac{2 \sqrt{6}}{3}=\frac{4 \sqrt{6}}{3} \end{aligned}
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