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?
On
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 $\cos 30^{\circ}.$
$$ h = \dfrac23\cdot \dfrac12 \cdot\dfrac{\sqrt 3}{2}=\dfrac{1}{2\sqrt 3}$$
$$ Area = \dfrac12 \cdot\dfrac{\sqrt 3}{2}=\dfrac{\sqrt 3}{4}$$
$$ Vol= Area\cdot 2 \pi h = \dfrac{\pi}{4}. $$
What kind of course is this for? I would think of this as a Calculus problem but you seem to be thinking there should be some geometry formula for this problem. We are given an isosceles equation in which the two congruent sides has length 1 cm and base angle 30 degrees. Dropping an altitude from vertex B to base AC we have two right angles with hypotenuses of length 1 cm and one angle 30 degrees. $cos(A)= \frac{\sqrt{3}}{2}= x/1= x$ where x is half the length of the base. The length of the base is $\sqrt{3}$. By the Pythagorean theorem the altitude of the triangle, h, is given by $h^2+ \frac{3}{4}= 1$ so $h= \frac{1}{2}$. That's the geometry, now the Calculus!
First set up a coordinate system in which the x- axis lies along the base, AC, with the origin, (0, 0), at the center of the base. Rotating the triangle around the base, AC, there are two ways to calculate the volume, the "method of disks" and the "method of shells". To use the method of disk we imagine, at each x, a line from (x, 0) up to a side of the triangle being rotated about x-axis. It sweeps out a disk of radius y so area $\pi y^2$. Taking the thickness of each disk to be "$\Delta x$", Each disk has volume $pi y^2 \Delta x$. Taking the limits as the number of disks goes to infinity, this becomes the integral [tex]\int \pi y(X)^2 dx.
So what is y(x) and what are the limits of integration? The two sides (other than the base, are lines from the base vertices, $\left(-\frac{\sqrt{3}}{2}, 0\right)$ and $\left(\frac{\sqrt{3}}{2}, 0\right)$ to the vertex $\left(0, \frac{1}{2}\right)$.
We can simplify this by recognizing the symmetry. We can calculate the volume for x> 0 and multiply by 2. For x> 0, y lies on the line from $\left(0, \frac{1}{2}\right)$ to $\left(\frac{\sqrt{3}}{2}, 0\right)$. Any non-vertical line can be written as $y= ax+ b$ for some constants a and b. Since $\left(0, \frac{1}{2}\right)$ is on the line we must have $\frac{1}{2}= a(0)+ b= b$. Since $\left(\frac{\sqrt{3}}{2}, 0\right)$ is on the line we must have $0= a\frac{\sqrt{3}}{2}+ b= a\frac{\sqrt{3}}{2}+ \frac{1}{2}$ so $a= \frac{\sqrt{3}}{3}$.
So $y(x)= \frac{\sqrt{3}}{3}x+ \frac{1}{2}$ and the integral is from 0 to $\frac{\sqrt{3}}{2}$. That is the volume is $2\pi \int_0^{\frac{\sqrt{3}}{2}}\left(\frac{\sqrt{3}}{3}x+ \frac{1}{2}\right)^2 dx$.