Find the volume of the solid if the cross-sections perpendicular to the x-axis are equilateral triangles. The base of a solid is in the region between the parabolas $x = y^2$ and $2y^2 = 3 - x$. Find the volume of the solid if the cross-sections perpendicular to the x-axis are equilateral triangles.
I need some help on this, kind of forgot my older calc. Thanks in advance!
 A: The geometry of this is simple.  One parabola faces out from the $y$ axis with vertex at the origin, the other has vertex at $x=3$ and faces opposite.  The parabolae intersect at $x=1$, so the cross-sections are bounded by the first parabola for $x \in [0,1]$ and the other for $x \in [1,3]$.
The area of an equilateral triangle of side $s$ is $\sqrt{3} s^2/4$.  When bounded within the curves, the side of a triangle at $x$ is $2 y(x)$.  The volume element is then $\sqrt{3} ([2 y(x)]^2/4) = \sqrt{3} y^2$. The volume is then
$$\sqrt{3} \int_0^3 dx \: y(x)^2 = \sqrt{3} \int_0^1 dx \: x + \sqrt{3} \int_1^3 dx \: \frac{1}{2} (3-x)$$
I trust you can evaluate these.
A: Make a sketch of the region of the $x$-$y$ plane described in the problem. I doubt you can solve te problem without a sketch. I certainly couldn't. Spend a little time visualizing the solid. It has a flat base that we have sketched. At the top, it has a sharp spine one wouldn't want to sit on.
The two parabolas meet at $x=1$, $y=\pm 1$. The parabola $2y^2=3-x$ meets the $x$-axis at $x=3$. 
By a formula that is undoubtedly familiar to you, the volume is the integral, from the beginning to the end,  of $A(x)\,dx$, where $A(x)$ is the area of cross-section at $x$.
To find this, recall that cross-sections are triangles. For some $x$, draw the vertical (parallel to the $y$-axis) line that passes throgh $(x,0)$. This meets one of the parabolas at two points. The line segment we get is the base of the triangle of cross-section. The solid itself sticks up from the paper. 
Note that there is a change at $x=1$.  The bounding curve from $0$ to $1$ is the curve $y^2=x$. From $1$ to $3$, the bounding curve is $2y^2=3-x$.
So we will have to break up our computation into two parts: from $0$ to $1$, and from $1$ to $3$.
Let us do the part from $0$ to $1$, or at least start on it.
For $0\le x\le 1$, the triangular cross-section has side $2y$, where $y=\sqrt{x}$.
What is the area of an equilateral triangle of side $s$? A bit of fooling around will show you it is $\frac{\sqrt{3}}{4}s^2$. For this you can use the Pythagorean Theorem, or trigonometric stuff you know about the $30$-$60$-$90$ triangle.
It follows that for $0\le x\le 1$, we have 
$$A(x)=\frac{\sqrt{3}}{4}(2y)^2=\sqrt{3}x.$$
Now you can integrate, quickly finishing the computation in the part from $x=0$ to $x=1$. The part from $1$ to $3$ is only a little more complicated, and is left to you.  
