# Find the slant area of a cone

Question: Find the slant curved area of the surface of revolution of a cone of semi-vertical angle $$\alpha$$ and base circle of radius a by revolving about the X-axis.

I tried using $$r=a \csc \theta$$ and integrating from $$\theta=0$$ to $$\theta=\alpha$$, but the answer is wrong.

Help me with the correct equation and the limits.

You should be careful to look to what you are integrating.

Consider a differential triangular area in yellow color on the slant side as shown:

$$dA= \frac12 \frac{a}{\sin \alpha} a \, d \theta$$

Integrating

$$\int dA = \int_0^{2 \pi} \frac12 \frac{a}{\sin \alpha} a \, d \theta = \frac{\pi a^2}{\sin \alpha}.$$

Also see how the standard slant area formula $$A = \pi a L$$ is derived with $$L= \dfrac{a}{\sin \alpha}.$$

HINT

Whe need to use the general formula for the area of a surface of revolution

$$S= 2\pi\cdot\int_a^b f(x)\cdot \sqrt{1+(f'(x))^2}dx$$

that is by

• $$f(x)=\tan \alpha \cdot x \implies f'(x)=\tan \alpha$$
• $$a=0 \quad b=a\cot \alpha$$

$$S= 2\pi\cdot\int_0^{a\cot \alpha} \tan \alpha \cdot x\cdot \sqrt{1+\tan^2 \alpha}dx$$

As an alternative and very effective way is use Pappus's Centroid Theorem.

• Yes, but I am not sure the function I used was correct. – Vishal Dalwadi Oct 28 '18 at 12:31
• @VishalDalwadi There was a typo in the main formula which I've fixed! I've also added some more detail for the function to be used and the set up of the integral by the standard method for surface of revolution. – user Oct 29 '18 at 7:34