Question

Area enclosed by curve $y^3-9y+x=0$ and Y - axis is -

• A. $\frac{9}{2}$
• B. $9$
• C. $\frac{81}{2}$
• D. $81$

Answer 1

The correct option is C $\frac{81}{2}$

The given equation of curve can be written as $x=f\left(y\right)=9y-y^3$. Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
$f\left(y\right)=y(9-y^2)$
$f\left(y\right)=y.(3+y)(3-y)$
So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0 f(y) is negative.
& from y = 0 to y = 3 f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
$A={\int }_{-3}^0|f\left(y\right)|dy+{\int }_0^{3}f\left(y\right)dyA={\int }_{-3}^0|9y-y^3|dy+{\int }_0^3(9y-y^3)dyA=-{\int }_{-3}^0(9y-y^3)dy+{\int }_0^{3}9y-y^{3}dyA=-[\frac{9y^2}{2}-\frac{y^4}{4}{]}_{-3}^0+[\frac{9y^2}{2}-\frac{y^4}{4}{]}_0^{3}A=-[0-0-\frac{81}{2}+\frac{81}{4}]+[\frac{81}{2}-\frac{81}{4}]A=81-\frac{81}{2}A=\frac{81}{2}$