Use a surface integral to show that the surface area of a right circular cone of radius $R$ and height $h$ is $\pi R \sqrt{h^2+R^2}$. Hint -- Use the parametrization $x=r\cos\theta$, $y=r\sin\theta$, $z=\dfrac{h}{R}r$, for $0\leq r \leq R$, and $0\leq \theta \leq 2\pi$.
Could someone explain how to set up this integral? What would you integrate?
This was the solution given, but I don't understand the steps.
Thanks.
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$\begingroup$Consider your cone lying in a xyz-space. Put the vertex of the cone at the origin, and imagine your base of cone lying in the plane $z=h$. Let $\alpha$ be apex angle so that $\tan\alpha=\cfrac{R}{h}$.
At height $z$ above the origin(vertex of cone), the radius of the base is $z\tan\alpha$. So lateral surface of the cone is represented by the region in $\mathbb{R}^3$, $R=\{(x,y,z):x^2+y^2=(z\tan\alpha)^2, 0\le z\le h\}$. To do the surface integral, we need to parameterize this, as follows:
$\gamma(r,\theta)=(r\cos\theta,r\sin\theta,\cfrac{r}{\tan\alpha})$, $r\in [0,h\tan\alpha]$, $\theta\in [0,2\pi)$
(the parameterization question asked for)
Now we find:
$\cfrac{\partial{\gamma}}{\partial{r}}=(\cos\theta,\sin\theta,\cfrac{1}{\tan\alpha})$
$\cfrac{\partial{\gamma}}{\partial{\theta}}=(-r\sin\theta,r\cos\theta,0)$
Then $\left|\cfrac{\partial{\gamma}}{\partial{r}}\wedge\cfrac{\partial{\gamma}}{\partial{\theta}}\right|=\left|\big(\cfrac{-r\cos\theta}{\tan\alpha},\cfrac{-r\sin\theta}{\tan\alpha},r\big)\right|=\cfrac{r\sec\alpha}{\tan\alpha}$.
Now do the surface integral:
$\iint\limits_R \,dx\,dy=\int\limits_{0}^{2\pi}\int\limits_{0}^{h\tan\alpha}\cfrac{r\sec\alpha}{\tan\alpha} \,dr \,d\theta=2\pi \cfrac{\sec\alpha}{\tan\alpha} \left[\cfrac{r^2}{2}\right]_0^{h\tan\alpha}=\pi h^2 \sec\alpha \tan\alpha=\pi h^2 \cfrac{R}{h} \cfrac{\sqrt{h^2+R^2}}{h}=\pi R\sqrt{h^2+R^2}$
$\endgroup$ 1 $\begingroup$I wouldn't use the hinted parametrization; just define a linear function which rotated on the x-axis generates your cone and then use the formular for lateral surfaces + the missing circlearea if needed:
$$M=2\pi \cdot \int_a^b f(x)\sqrt{1+(f^{\prime} (x))^2}\, \mathrm dx$$
Where $f(x)= \frac Rh x$ or $f(x)= -\frac Rhx+R$
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