A paraboloid has equation $z=a-x^2-y^2$ and a plane the equation $z=\lambda a$, where $0< \lambda < 1$. $V(A)$ is the volume of the paraboloid between its vertex and the given plane. $V(B)$ is the volume of the paraboloid between the given plane and the plane $z=0$. Find $\lambda$ such that $V(A)=kV(B)$
In cylindrical coordinates, for $V(A)$, $z$ goes from the plane $z=\lambda a$ to the paraboloid $z=a-x^2-y^2$. You get the radius equating $\lambda a=a-x^2-y^2$
The resulting integral is $$\int_0^{2\pi} \int_0^{\sqrt {a(1- \lambda)}} \int_{ \lambda a}^{a-r^2}\ r\ dz\ dr\ d \theta = \frac{a^2(1- \lambda^2) \pi}{2}$$
For V(B) I didn't understand very well how to find the limits of integration. I tried for $z=a-x^2-y^2$ to the plane $\lambda a$. I found the projection of this solid on the $z=0$ plane equating $0=a-x^2-y^2$. And I got a radius of $\sqrt a$. The integral for this volume is:
$$\int_0^{2\pi} \int_{0}^{\sqrt a} \int_{ a-r^2}^{\lambda a}\ r\ dz\ dr\ d \theta = \frac{a^2(2 \lambda -1) \pi}{2}$$
To find $\lambda$ I use $V(A)=kV(B)$
$$\frac{a^2(1- \lambda^2) \pi}{2}=\ k \frac{a^2(2 \lambda -1) \pi}{2}$$
I solved for $\lambda$ in the resulting equation $- \lambda^2 - 2k \lambda +(k+1)=0$
$\lambda =\ -k \mp \sqrt{k^2+k+1}$
The answer isn't right.