Question
The area bounded by the curve $${x^2} = ky,\,k > 0$$ and the line $$y=3$$ is $$12\,{\text{uni}}{{\text{t}}^2}.$$ Then $$k$$ is :
A.
3
B.
$$3\sqrt 3 $$
C.
$$\frac{3}{4}$$
D.
none of these
Answer :
3
Solution :
$$\eqalign{ & {\text{Area}} = 2\int_0^3 {x\,dy} \cr & = 2\int_0^3 {\sqrt {ky} } \,dy \cr & = \left[ {2\sqrt k .\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^3 \cr & = \frac{{4\sqrt k }}{3}.3\sqrt 3 \cr & \therefore 12 = 4\sqrt k .\sqrt 3 \,\,\,\,\, \Rightarrow k = 3 \cr} $$
$$\eqalign{ & {\text{Area}} = 2\int_0^3 {x\,dy} \cr & = 2\int_0^3 {\sqrt {ky} } \,dy \cr & = \left[ {2\sqrt k .\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^3 \cr & = \frac{{4\sqrt k }}{3}.3\sqrt 3 \cr & \therefore 12 = 4\sqrt k .\sqrt 3 \,\,\,\,\, \Rightarrow k = 3 \cr} $$