1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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:

enter image description here

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

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}. $

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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

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

enter image description here

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, but I am not sure the function I used was correct. $\endgroup$ – Vishal Dalwadi Oct 28 '18 at 12:31
  • $\begingroup$ @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. $\endgroup$ – user Oct 29 '18 at 7:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.