Question

find the volume of the solid obtained by rotating the region bounded by
y=(x^2)/4
x=2
y=0
about the y-axis

Answer 1

The whole cyllindrical solid is \(\pi\cdot 2^{2}\cdot 1 = 4\pi\). What's the interior part that you must discard?

Answer 2

just that was the easy part . . . I will try to explain where I need help
my math professor was a little confusing, I just do not full understand how to set up the integral, I drew a picture and everything but just cannot put the pieces together. Once I have the integral I can calculate it fine

Answer 3

so would it be
\[\pi \int\limits_{0}^{1}2-2\sqrt{y} dy\]

Answer 4

Too bad this is not an evaluate-the-integral problem. It is a set-up-the-integral problem.

Think on this formula: Volume of Right Circular Solid: A = \(\pi r^{2}h\)

\(\dfrac{\partial A}{\partial r} = 2\pi rh\)

\(\dfrac{\partial A}{\partial h} = \pi r^{2}\)

Answer 5

This is why I find this teacher so confusing he does one very basic general diagram and then goes striaght into exampes . . . he never even mentioned the formula you are talking about, I think (he also has a very heavy Russian accent)

Answer 6

can you tell me what specific part of my integral is wrong so that I know where to look?

Answer 7

No one tends to mention that relatioship. It's there, though.

Answer 8

Why do you think there is anything wrong with your integral? Can you do it the other way?

Answer 9

I just get confused when it is rotated about the y axis and sometimes mix up my limits of integration and the outside function also I didn't have the integral when I posted the question it just dawned on me when I posted it.

Answer 10

Well, since you are using Disks, there should be a single argument and it should be squared. Please refer to my second mysterious formula.

Note: \(\pi\int\limits_{0}^{1}2\;dy = 2\pi\). That is NOT \(4\pi\) as would be correct for the entire cyllindrical solid. However, \(\pi\int\limits_{0}^{1}2^{2}\;dy = 4\pi\). Now THAT show promise.

If you ponder my mysterious formulas, you will see it and you will not ever have to be confused.

Answer 11

Seriously, just square your arguments

\(2^{2}\) and \((2\sqrt{y})^{2}\)