Blog #23

Cross-Sections:

Ex. 1) Find the volume of a solid with triangular perpendicular cross-section for: y = 1 – x/2 & y = -1 + x/2.
To solve a cross-section problem, you simply do what we’ve been doing with area and volume this whole time, just with the area formula for whatever shape they give you. In this case, that would be a triangle.
The area of a triangle: ½ bh
So, you simply take the integral of base times height (the first equation times the second equation) and multiply it by ½. Oh, and don’t forget the dx.
½ S(1 – x/2)(-1 + x/2)dx = ½ S(-1 - x²/4 + x)dx
For this particular problem, they did not give you any bounds to work with, so you just leave the answer as an equation.

For other problems like this but with different shapes, use these area formulas:
Triangles >>> S ½ bh
Rectangles >>> S lw
Semi-cirlces >>> S ½ π r²
Cirles >>> S π r²
Squares >>> S s²

Ex. 2) Find the volume of a solid with square perpendicular cross-section for: y = x² & y = x + 1.
Given that our shape is a square, we are going to use S s². Because of the two equations, we are first going to subtract the two equations, then square it.
S ((x+1) – (x²))² dx
Be sure to put the graph that is one top first.
S (x + 1 – x²)² dx
Since we are not given any bounds, the answer stays as an equation.