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Mathematics · Area under the Curves

JEE Main 2026 — 22 January, Evening Shift — Question 12

The area of the region A={(x,y):4x2+y2≤8\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}): 4 \mathrm{x}^{2}+\mathrm{y}^{2} \leq 8\right. and y2≤4x}\left.\mathrm{y}^{2} \leq 4 \mathrm{x}\right\} is :

  1. Option A:

    π2+2\frac{\pi}{2}+2

  2. Option B:

    π+23\pi+\frac{2}{3}

    Correct
  3. Option C:

    π+4\pi+4

  4. Option D:

    π2+13\frac{\pi}{2}+\frac{1}{3}

Answer: B

Step-by-step solution

A=∫022xdx+2∫128−4x2dx\mathrm{A}=\int_{0}^{2} 2 \sqrt{\mathrm{x}} \mathrm{dx}+2 \int_{1}^{\sqrt{2}} \sqrt{8-4 \mathrm{x}^{2}} \mathrm{dx}

=83(x32)∫01+4∫122−x2dx=\frac{8}{3}\left(x^{\frac{3}{2}}\right) \int_{0}^{1}+4 \int_{1}^{\sqrt{2}} \sqrt{2-x^{2}} d x

=83+4×12[x2−x2+2sin⁡−1(x2)]∣12=\frac{8}{3}+4 \times\left.\frac{1}{2}\left[x \sqrt{2-x^{2}}+2 \sin ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right]\right|_{1} ^{\sqrt{2}}

=83+2[2×π2−1−2×π4]=\frac{8}{3}+2\left[2 \times \frac{\pi}{2}-1-2 \times \frac{\pi}{4}\right]

=83+2π−2−π=π+23=\frac{8}{3}+2 \pi-2-\pi=\pi+\frac{2}{3} sq. units

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Area under the Curves
Topic
Area under the Curves