The area bounded by the curves $x=4-y^2$ and the $Y$-axis is

• A

  16 sq. unit
• Correct answer
  B

  $\frac{32}{3}$ sq. unit
• C

  $\frac{16}{3}$ sq. unit
• D

  32 sq. unit

Solution

Hint: $x=4-y^2 \Rightarrow y^2=-x+4$ $x=0 \Rightarrow y^2=4 \Rightarrow y= \pm 2$ Area bounded by the curve $x=4-y^2$ and $y$-axis is, $=2 \int_0^2 x d y=2 \int_0^2\left(4-y^2\right) d y=2\left[4 y-\frac{y^3}{3}\right]_0^2=2\left(8-\frac{8}{3}\right)=\frac{32}{3}$ sq. unit

Hence, the correct answer is option: 'B'.