Todays Scribe Post: Cross-Sections.

Before we started with our lesson, we talked about some other things. For example we talked about how Obama wrote a book on how he will reform the education system in the United States of America. I wouldn't mind reading this, but then again I wouldn't have to know about this due to us being Canadian and all. This discussion led to the awesome education system of Canada, and everything awesome about Canada. Except for our phone companies, which are pretty much trash.

All the randomness aside, we actually continued with our lesson on applications of integrals, we learned something new. We were supposed to learn this before rotations of 2d graphs and so forth, but we did not. We pretty much learned the hard stuff first that mainly dealt with the rotation of 2d objects which would make up a 3d object, which we had to visualize. This is cross-sections not rotations. A good example of cross-sections is like slices of bread from a bread loaf, and my favorite example is sushi sliced from a sushi roll. It looks like a rectangle from Birdseye view but when sliced, each cross-section is a circle! Amazing?! I know, just like magic.

Our first problem looked like this:




As the question states, we have to find the volume of the 3d shape the graph makes between the graph itself, the x-axis, x=0 and x=2. The formula used is the formula for the area of a circle, as shown in the slide. The only dimension we needed was the radius, and that's already given by the graph which is x². So we input this into the equation, and integrate it between the interval 0 to 2. Easy enough.

For our 2nd slide we had something a bit harder, but pretty much the same concept, but now we're dealing with different shapes than circles, such as semi-circles, equilateral triangles and squares, so our area equations would be different for each one.





The width of each cross-section would be dx in the integral equation. The length or radius would depend on the given function of the graph, in this case we have x².
Now the length of each cross-section when considering the semi-circle would be the diameter, and we would divide this by 2 to get the radius, r = x²/2. The equation of the area would be A = pi*r²/2 and we input r = x²/2 for the radius, and solve over the given interval, simple enough.
The square would be easy solve as the equation for the area is A = s² and s = x².
The equilateral triangle would have a base of x² and the height would have to be found using a bit of calculations. Imagine cutting the triangle in 2 equal parts right down the middle, splitting the base in two. The line used to cut would be the height. This would make a right angle triangle, and knowing an equilateral triangle all the angles are pi/3. Using tan(pi/3) = h/(b/2) we find that height is:
The area of a triangle is A = bh/2. We know b = x² and h. Throw all that together and we get this:We integrate it over the interval 0 to 2, and voila. As long as we know the shape of the cross-section, the corresponding area equation for this shape and the values for the dimensions to input into the equation then we know the anwser, its simple as pie. Pi day is almost here. Our homework was a worksheet handed to us entitled "Volumes of Revolution Excerises" and this should be worked on for continued success. The next scribe will be Lawrence.