Let $ABC$ be an isosceles triangle with $AB=BC=1$ cm and angle $A=30$ degrees. Find the volume of the solid obtained by revolving the triangle about the line $AB$.
I don't know how to start this problem because it seems like it wouldn't be a cone. Can someone help with the first few steps or give a sketch on how to go about this?
Best Answer
Not sure if integration can be used. Volume is computed by Pappu's thm, an easier way out. Triangle ABC is sketched to proportion.
Red line radius $h$ drawn upto center of gravity of triangle ABC from axis of revolution AB. It is two-thirds of triangle altitude multiplied by $\cos 30^{\circ}.$
$$ h = \dfrac23\cdot \dfrac12 \cdot\dfrac{\sqrt 3}{2}=\dfrac{1}{2\sqrt 3}$$
$$ Area = \dfrac12 \cdot\dfrac{\sqrt 3}{2}=\dfrac{\sqrt 3}{4}$$
$$ Vol= Area\cdot 2 \pi h = \dfrac{\pi}{4}. $$