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.
Answer
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}. $
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