Question

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

• A. $\frac{3}{32}$
• B. $\frac{16}{4}$
• C. $\frac{32}{3}$
• D. $\frac{33}{2}$

Answer 1

The correct option is C $\frac{32}{3}$

The given curve is $x=4-y^2\cdots \left(i\right)$
$\Rightarrow y^2=-(x-4)$
This is the equation of a left handed parabola with vertex $A(4,0)$
Putting $x=0$ in $\left(i\right)$ we get: $y=\pm 2$
$\therefore$ The parabola intersects $y-$axis at points $B(0,2)$ and $C(0,-2)$


Required area $=$ Area of shaded region $ABCA$
$=2\times$ Area of region $AOBA$
$[\because$ The region is symmetrical about $x-$axis$]$
$=2\underset{0}{\overset{2}{\int }}xdy$
$=2\underset{0}{\overset{2}{\int }}(4-y^2)dy=2{[4y-\frac{y^3}{3}]}_0^2$
$=\frac{32}{3}$ sq. units