Solution 8: Consider the graph of $ \ y = \displaystyle{r \over h} x \ $ on the interval $ [0, h]$. Form a cone of height $h$ and base radius $r$ by revolving this graph about the $x$-axis. Now find its surface area. Here is a carefully labeled sketch of the graph with a radius $r$ marked together with $x$ on the $x$-axis.
Thus the total Area of this Surface of Revolution is
$$ Surface \ Area = 2 \pi \int_{0}^{h} (radius) \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx $$
$$ = 2 \pi \int_{0}^{h} \Big( {r \over h} x \Big) \sqrt{ 1 + \Big( { r \over h } \Big)^2 } \ dx $$
$$ = 2 \pi \int_{0}^{h} \Big( {r \over h} x \Big) \sqrt{ 1 + { r^2 \over h^2 } } \ dx $$
$$ = 2 \pi \int_{0}^{h} \Big( {r \over h} x \Big) \sqrt{ { h^2 \over h^2 } + { r^2 \over h^2 } } \ dx $$
$$ = 2 \pi \int_{0}^{h} \Big( {r \over h} x \Big) \sqrt{ h^2 + r^2 \over h^2 } \ dx $$
$$ = 2 \pi \int_{0}^{h} \Big( {r \over h} x \Big) { \sqrt{ h^2 + r^2 } \over \sqrt{h^2} } \ dx $$
$$ = 2 \pi \int_{0}^{h} \Big( {r \over h} x \Big) { \sqrt{ h^2 + r^2 } \over h } \ dx $$
$$ = 2 \pi {r \over h^2} \int_{0}^{h} x \cdot \sqrt{ h^2 + r^2 } \ dx $$
$$ = 2\pi \cdot {r \over h^2} \cdot \sqrt{ h^2 + r^2 } \cdot {1 \over 2} x^2 \Big|_{0}^{h} $$
$$ = 2\pi \cdot {r \over h^2} \cdot \sqrt{ h^2 + r^2 } \cdot {1 \over 2} ( (h)^2 - (0)^2 ) $$
$$ = \pi r \sqrt{ h^2 + r^2 } $$
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