The nearly forgotten blog post: Solids

Hey guise, its *squeak* Not Paul here, blogging about our class.

The nearly forgotten part is because its like 1AM and I was about to call it a night when I booted up my browser to check my email. What pops up? The blog. And I'm like, where's the scribe?

OH RIGHT!

So yeah, here we go again.

Today we worked on the stuff we sort of got introduced to yesterday, which was Solids. How do you find the area between two functions?! THATS CRAZY MAN.

In reality, it ain't so crazy, man. thinking about it logically, the integral can be found by integrating over the interval that ranges from the first intercept of the two functions to the second intercept (they NEED to have two intercepts, oth

erwise it'd go on forever).

I think this slide describes better what I'm tryin

g to say. Because me English so good not as much.

[Click pics for a larger version]

What's going on here is that we have two functions: f(x) and g(x). f(x) is equal to root x and g(x) is equal to x squared. Our mission, should we choose to accept it, is to find the area between the two functions. To do this, first we find the intercepts. The first one we know is zero, just by looking at the graph/sketch, or by simple reasoning. The second one can be found by using the equation f(x) = g(x) and solving for the variable (in this case x). This is because we want to find the value where f(x) and g(x) are equal. Doing this yields us the value 1 for the second intercept. Yay!

Now, we create an equation to find the area. Subtract one integral from the other, but both must be on the same interval (intercept 1 to intercept 2). Antidifferentiate, solve with fundamental theorem of calculus! Bada bing.

Now we move onto something a little trickier. Our first function, f(x), is e to the power x. Funny how that little guy seems to pop up everywhere. The second function, g(x), is 4 minus x squared.

Now, I said a little trickier, but really, is it any harder? No! The only difference is it might take a little longer due to anti differentiation, but longer is not equal to harder! You can even speed things up by graphing and using your graphing calculator to find the intercepts.  Do a little antidifferentiation... algebra, and you're done before you know it.

Eaaaaaaaaasy.

 

Bread: designed to make us hungry. *eats toast*

But, also a cool reminder that in math that it’s okay to approximate just a little. I mean, nothing is really truly exact. So what if that slice of bread really has that little bit of z axis breadth (pun intended)? Treat it like a 2 dimensional object, relatively, its no big deal.

 

 

 

 

Now here’s when we get into the good stuff. Like shapes. Here, we use the circle area equation to find the integral of this cone. However, the cone is mapped to a 2D plane, so you cant really see all of it, but rest assured, its a cone. Now, see, this cone has the property that its radius at a certain height can be found using the equation r = –(1/2)x + 4. Do we really need any more information? We have the interval to integrate from (0 to 8, because the height is 8), and the function to integrate (area). Solution in the next slide:

 

 

And finally, a general rule to guide us. This should come in handy…

 

And with that, Im am done and tired, and in desperate need of some sleeeeeep.

 

Next scribe is the one and only Kristina.

 

Night guys.