Calc 2 HW help!! (Integrals)?

Question

I have an application problem that asks me to solve for the volume of a bowl.

"A bowl has a shape that can be generated by revolving the graph of y=(x^2)/2 between y=5 and y=0 about the y-axis. Find the volume of the bowl."

The formula I need to use is V= ∫ A(x) dx

I figured since it was rotated around the y-axis, i'd have to use the disk method. But that would require me to solve for the area of the parabola y=(x^2)/2 from y=0 to y=5. I remember that also had something to do with integrals, but I did that last year, and I don't remember!

Is there any other way to do it?

Answer 1

method of disks is actually very easy here.  draw a cross section.  for any value of y, the radius of the disk is x, the height is dy, and so the volume is πx² = π(2y).

so just work out integral(0 to 5) of 2πy dy and get πy² (evaluated at 5 and 0), so V = π•5² = 25π

in general, if the radius of the paraboloid at height h is a, the volume is ½πa²h.

Answer 2

I don't follow why you say you "need" to use the formula

V= ∫ A(x) dx.

In this problem it is more natural to do an integral over y.  You should imagine the volume as a stack of thin plates, starting from zero radius at the bottom where y = 0 and growing larger as y rises to 5.

The radius of each plate is the value of x at a given height y, so you need to solve the boundary formula y=(x^2)/2 for x --

x = sqrt(2y)

Now we can write the formula for the volume of a circular disk of thickness dy at height y.  It's just the area of a circle of radius sqrt(2y) times thickness dy:

pi r^2 dy = pi [sqrt(2y)]^2 dy = pi 2y dy

Then the volume is simply all the disks from y = 0 to 5 added up:

V= ∫ pi 2y dy  [y = 0 .. 5]

= pi y^2 [0 .. 5]

= 25 pi