1. **State the problem:** We have a sphere and a cone with the same curved surface area. The sphere has radius $3y$ and the cone has radius $3y$ and slant height $l$. We need to: a) Show that the surface area of the sphere is $36\pi y^2$. b) Find an expression for $l$ in terms of $y$ given that the curved surface area of the cone equals the surface area of the sphere. 2. **Recall formulas:** - Surface area of a sphere: $$4\pi r^2$$ - Curved surface area of a cone: $$\pi r l$$ 3. **Part (a): Surface area of the sphere** Given radius of sphere $r = 3y$, substitute into the sphere surface area formula: $$4\pi (3y)^2 = 4\pi (9y^2) = 36\pi y^2$$ This shows the surface area of the sphere is $36\pi y^2$. 4. **Part (b): Expression for $l$ in terms of $y$** The curved surface area of the cone equals the surface area of the sphere: $$\pi r l = 36\pi y^2$$ Substitute $r = 3y$: $$\pi (3y) l = 36\pi y^2$$ Simplify by dividing both sides by $\pi$: $$3y l = 36 y^2$$ Divide both sides by $3y$ (assuming $y \neq 0$): $$l = \frac{36 y^2}{3 y} = 12 y$$ **Final answers:** - Surface area of the sphere is $36\pi y^2$. - Slant height of the cone is $l = 12 y$.