What is the measure of the vertex angle of the cone made of a circular sector of angle $270^\circ$?
I worked out the following solution by my own, but I get stuck.. Can someone find my mistake?
Let the (semi) vertex angle of the cone be $\alpha$ ($0\leq \alpha\leq\frac{1}{2}\pi$) and the radius $r$. Then the surface area of the solid of revolution formed by revolving $y=x\tan\alpha$ around the $x$-axis between $x=0$ and $x=\frac{r}{\tan\alpha}$ is
$$2\pi\int_0^{\frac{r}{\tan\alpha}}x\tan\alpha\sqrt{1+\tan^2 \alpha}\,dx=2\pi\tan\alpha\sec\alpha\int_0^{\frac{r}{\tan\alpha}}x\,dx=\frac{\pi r^2}{\sin\alpha}$$
The surface area of the circular sector is $\frac{3}{4}\pi r^2$ and so these two have to be equal to each other
$$\frac{\pi r^2}{\sin\alpha}=\frac{3}{4}\pi r^2 \iff \sin\alpha=\frac{4}{3}$$
There seems to be no solution for $\alpha$..
Best Answer
Let the radius of the circular sector be $R$. Then the curved length subtended by the $270^{\circ}$ angle of this sector will be $2\pi R\times\frac{270}{360}=\frac{3\pi R}{2}$.
This length must equal the circumference of the base of the cone. So, if we let the radius of the base of the cone be $r$, then we have:$$2\pi r=\frac{3\pi R}{2}$$$$\therefore r=\frac{3R}{4}$$We also know that the length of the side of this cone must be $R$. So, if the semi vertex angle of the cone is $\alpha$, then we get:$$\sin(\alpha)=\frac{r}{R}=\frac{3}{4}$$ Hopefully you can solve from here.
Using your integral method, the limits of the integral should be from $0$ to $r\cos(\alpha)$ and it should be of the form:$$SA=\int_0^{r\cos(\alpha)}2\pi x\tan(\alpha).\frac{dx}{\cos(\alpha)}$$ If you work this through then you will get the correct answer.
For a more detailed description of this method see Setting Up an Integral to Find A Cone's Surface Area