The surface area of sphere with $r$ that is made by play dough is $A_1$. The surface area of cylinder is being $A_2$ when that's made by this sphere. What is the ratio of $\frac {A_1}{A_2}$?
I've tried to get what is meant in question.
$$A_1 = A_2$$
By the way
$$4 \pi r^2 = 2\pi r h + 2\pi r^2$$
and
$$2\pi r^2 = 2 \pi r h$$
According to me, the surface area of sphere must be equal to the surface area of cylinder.
The answer seems $\frac {6}{7}$ on my textbook. How did we arrive at that answer?
Might I get your hints?
The volume of the sphere is equal to the volume of the cylinder, so $$\frac{4}{3}\pi r^3 = \pi r^2 h\,.$$ Rearrange this to get $$h=\frac{4}{3}r.$$ Thus $$\frac{A_1}{A_2}=\frac{4\pi r^2}{2\pi r^2+2\pi rh}.$$ Substituting $h=\frac{4}{3}r$, you will find $\frac{A_1}{A_2}=\frac{6}{7}$.