Mystery shapes - making shapes from vague descriptions I am suppose to mentally create something that has base $x^2 + y^2 = 1$ and the cross sections perpendicular to the x axis are triangles whose height and base are equal.
What is going on? I tried to graph it but it was too hard, is this like a turtle? Sphere on back, pointy spikes on the bottom? What does "triangles" mean? How many? 2?
 A: This screenshot from Calculus Early Transcendentals (6th Ed.) by Stewart might help:

Note that the example is slightly different; each cross section has a base of $2y$ and a height of $\sqrt{3}y$, whereas in your example you want a base of $2y$ and a height of $2y$. So your $3$D diagram should be similar, except more vertically stretched.
A: Hint: you have a "cone" in 3D: and I'm assuming you can take the x-y plane as where the cone sits: its base is a circle of radius $1$. Think of an upside-down ice-cream cone with no ice-cream, whose height is equal to the diameter of the circle. Can you fill in more of the details? Try to envision how "steep" the cone must be from the circumference of the base of the cone, to its tip.

Added: The cross section will be an equilateral triangle: If we call of the base "2y", then 1/2 base will be y, but since we have that height will equal base, (given), we must have $2y$. cross sectional Area (area of triangle) will then be $$A = \frac 12\;\text{base}\, \times\, \text{height} = (y)(2y) = 2y^2.\tag{Area}$$ Now, we know that $x^2 + y^2 = 1$, so $y^2 = 1 - x^2 \implies 2y^2 = 2(1 - x^2)$. Substituting this into our equation for area, to get an equation $A(x)$ in terms of $x$ gives us $$A(x) = 2y^2 = 2(1-x^2) = 2 - 2x^2\tag{A(x)}$$ Now, we want to sum  the area of all cross sections (triangles) between $x = -1\;\text{and}\; x = 1$, to determine the volume $V$ of the cone:  
$$V = \int_{-1}^1 A(x)\, dx = \int_{-1}^1 (2-2x^2)\, dx$$
