Show that the surface area of a cone with base radius $a$ and height $h$ is $\pi a \sqrt{a^2+h^2}$
$Hint$: use the surface area formula, A(S) = $\iint_D \sqrt{(\frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2 +1 } \,dA$
I know the equation for a cone is $z^2 = x^2 +y^2$. But how do I bring $a$ and $h$ into this equation?
Hint: The perimeter of the circle $a = x^2 + y^2$ is $2 \pi a$. You want to integrate over $y$ from $0$ to $h$.
I started typing this before the suggestion to use the double integral appeared, anyway, here is the analytic solution.
First of all you need to derive the correct equation of the cone. In this case it will be
$$(z-h)^2 = \frac{h^2}{a^2} \left(x^2+y^2\right)$$
Consider a quarter of the cone in the first octant. The integrand is going to be
$$\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}=\frac{\sqrt{h^2+a^2}}{a}$$
Now the limits of integration are set as follows. $x$ varies from $0$ to $a$ while $y$ varies from $0$ to $\sqrt{a^2-x^2}$.
$$\frac{1}{4}=\frac{\sqrt{h^2+a^2}}{a}\int_0^a dy \int_a^{\sqrt{a^2-x^2}} dx\\=\frac{\sqrt{h^2+a^2}}{a}\int_0^a \sqrt{a^2-x^2} dx$$
The last integral equals $\frac{\pi a^2}{4}$ from which the result follows.
Here is a geometric derivation. If you cut the cone along a generating line and unfold you get a sector of, say $\phi$ radian and radius $r$. Length of the subtended arc is therefore $$L=\phi r$$ Area of the sector is $$S=\frac{1}{2}\phi r^2$$ Now we need to relate $\phi$ and $r$ to $a$ and $h$. When you fold the sector back into the cone $L$ becomes the circumference of the base with radius $a$. Hence $$2\pi a = \phi r$$ $$\phi = \frac{2\pi a}{r}$$ Now from the axial cross-section it is clear that $$r=\sqrt{a^2+h^2}$$ So finally $$S=\frac{1}{2}\frac{2\pi a}{\sqrt{a^2+h^2}}\left(a^2+h^2\right)=\pi a \sqrt{a^2+h^2}$$
