The diagrams below show a solid hemisphere and a solid cone. The hemisphere has a radius of 3y cm. The cone has a radius of 2y cm and slanted height l cm.

1. Write the total surface area of the solid cone as curved surface area plus base area.

For a solid cone,
$\text{Total surface area}=\pi rl+\pi r^2$
where $r$ is the radius and $l$ is the slant height.

2. Substitute $r=2y$ and simplify to the required form.

Given $r=2y$,
$\text{Total surface area}=\pi(2y)(l)+\pi(2y)^2$
$=2\pi yl+4\pi y^2$
$=2\pi y(l+2y)$
Hence, the total surface area of the solid cone is $2\pi y(l+2y)\text{ cm}^2$.

1. Find the total surface area of the solid hemisphere.

For a solid hemisphere,

$\text{Total surface area}=\text{curved surface area}+\text{circular base area}$

$=2\pi r^2+\pi r^2=3\pi r^2$

Given $r=3y$,

$\text{Total surface area}=3\pi(3y)^2=27\pi y^2$

2. Equate the hemisphere's total surface area to the cone's total surface area.

From part (a), the total surface area of the cone is

$2\pi y(l+2y)$

Since the two total surface areas are equal,

$27\pi y^2=2\pi y(l+2y)$

3. Solve the equation for $l$ in terms of $y$.

Since $y>0$, divide both sides by $\pi y$:

$27y=2(l+2y)$

$27y=2l+4y$

$2l=23y$

$l=\frac{23y}{2}$

1. Use the hemisphere volume formula to find $\pi y^3$.

For a hemisphere of radius $3y$,
$\text{Volume}=\frac{2}{3}\pi r^3$
$729=\frac{2}{3}\pi(3y)^3$
$729=18\pi y^3$
$\pi y^3=\frac{729}{18}=40.5$

2. Find the perpendicular height of the cone using the slant height from part (b).

From part (b), $l=\dfrac{23y}{2}$ and the cone radius is $2y$.

Let the perpendicular height of the cone be $h$.
$h^2=\left(\frac{23y}{2}\right)^2-(2y)^2$
$h^2=\frac{529y^2}{4}-4y^2$
$h^2=\frac{513y^2}{4}$
$h=\frac{\sqrt{513}}{2}y=\frac{3\sqrt{57}}{2}y$

3. Calculate the volume of the cone.

$\text{Volume of cone}=\frac{1}{3}\pi r^2h$
$=\frac{1}{3}\pi(2y)^2\left(\frac{3\sqrt{57}}{2}y\right)$
$=2\sqrt{57}\pi y^3$

Since $\pi y^3=40.5$,
$\text{Volume}=2\sqrt{57}(40.5)=81\sqrt{57}$