Deriving the volume of a sphere solution

note* I'm assuming you are already familiar with calculus. If you're not, it's probably not going to make any bit of sense.

I will be doing this using integration techniques. Remember learning about solids of revolution from calc. I? That's what I'll be using. I'm not really sure if it's the best way, but it works.

Consider the graph of a basic semi-circle. If you were to revolve it around the x-axis, the solid generated would be a sphere whose cross sections are circles.

Imagine slicing through a sphere. You'll notice that the cross sections are circles of varying sizes. What I'll essentially be doing is creating a summation of disks that each have a very small thickness of dx.

To start off, I'm going to create a function of the cross sectional area and integrate it from one end of the semi circle (-r) to the other (r). I'll need the formula for the area of a circle. The radius of each cross section will vary as a function of x, and so I need to make the appropriate substitution for the radius. Through integration, the disks of varying cross sectional area will be summed together to create a sphere.

There's nothing all that tricky about solving this integral. If you work it all out you should end up with the correct formula for the volume of a sphere.

Deriving the volume of a sphere

For this next one I want to derive the volume of a sphere using calculus. I don't have much more to add, but I feel like I should at least post a picture of something...

problem 6: the derivative of y=x^x^x

When I saw the solution to this, I thought it was really cool. Check it out at khanacadamy.org. This website really helped me get through calc. I.
https://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/calculus--derivative-of-x--x-x

problem 5: solution to the slightly harder integral

I must say, this was a bit more challenging, but nothing I can't do.

Using properties of logarithms, I can rewrite the integral like this:

It simplifies to this:

Now I just need a little u-substitution. 

For u-substitution to work, I'll also have to rewrite the integral as follows:

After doing u-substitution, it should look like this:

The integral of ln(x) requires integration by parts. I'm not going to go through the process now because I did it in a different post. It should like this this:

Well, it's simple from here. Just plug back in (x^2+1) in place of u. I'm going to leave it in terms of u because it's long and messy in terms of x.

problem 5: A slightly harder integral

My brother said this problem was on one of his Calc. II exams and challenged me to do it. Challenge accepted

Problem 4 solution: The ACT math problem.

Well, this is what we started out with. It doesn't look like there's much we can do, but the coordinates of point C can be found by dividing the figure into multiple 30-60-90 triangles.

 Based on the proportions of a 30-60-90 triangle, we can figure out the length of each side in terms of w. The shortest leg is w, the adjacent leg is w√3, and the hypotenuse is 2w.  Put it all together and you should get (w√3,2w).

problem 4: An ACT math problem

I was taking a practice test in my Princeton review book and I came across this problem. ACT math is usually pretty boring, but this one had an interesting solution.

Triangle ABC, shown above in the standard (x,y) coordinate plane, is equilateral with vertex A at (0,w) and vertex B on the x-axis as shown. What are the coordinates of vertex C?