The base of S is the triangular region with vertices (0,0) (1,0) and (0,1). Cross-sections perpendicular to the y-axis are equilateral triangles.
WHat's the volume of the solid?
(I have no idea!!!)
I'm confused:
a) (0, 0), (1, 0) and (0, 1) are the vertices of an isosceles right triangle. To get equilateral triangles as cross-sections the solid has to be distorted heavily.
b) the height of the solid isn't mentioned so the volume could be zero or unlimited ...
Please send the exact wording of your problem.
Thanks a lot for you help...
Unfortunately that is the exact wording of the problem. The base of the shape is the right triangle described in the problem. The height is described by an equilateral triangle that is perpendicular to the Y axis. I think i can visualize it, but I don't know if I made it a little bit clear.
Well then it is much easier to calculate the volume if you take the red equilateral triangle as base:
The height in an equilateral triangle with side 1 is (use Pythagorean theorem):
$\displaystyle h= \frac12 \cdot \sqrt{3}$
The base area then is:
$\displaystyle A= \frac12 \cdot h \cdot 1= \frac12 \cdot \frac12 \cdot \sqrt{3} \cdot 1= \frac14 \cdot \sqrt{3}$
The height H of the solid is the edge(?) form (0, 0) to (0, 1) that means H = 1
Use the formula of the volume of a pyramide:
$\displaystyle V_{pyr} = \frac13 \cdot A \cdot H$ . Plug in all values you know:
$\displaystyle V_{solid} = \frac13 \cdot \frac14 \cdot \sqrt{3} \cdot 1 = \frac1{12} \cdot \sqrt{3}$
EDIT: You probably have noticed that all the nice calculations are superfluous
1. The height in the equilateral triangle has the same length as the height of the solid.
2. The base area (painted blue) is half of a square with side 1.
3. Use the volume formula of the pyramide:
$\displaystyle V_{solid} = \frac13 \cdot \frac12 \cdot \frac12 \cdot \sqrt{3} = \frac1{12} \cdot \sqrt{3}$
According to your headline you are supposed to use integrals:
The length of the side of one cross-sectional equilateral triangle is (1-y). Therefore the area of such atringle is:
$\displaystyle A=\frac14 \cdot (1-y)^2 \cdot \sqrt{3}$
Consider the volume of the solid as a lot of thin slices of equilateral triangles. The thickness of one slice is dy. The volume is the sum of all possible slices:
$\displaystyle V=\int_1^0\left(\frac14 \cdot (1-y)^2 \cdot \sqrt{3}\right)dy = \left. \frac1{12} \cdot (1-y)^3 \cdot \sqrt{3}\right|_1^0 = \frac1{12} \cdot \sqrt{3} - 0$