Calculus: Integration?

There seems to be some confusion here. If the vase was all solid you would have trouble putting flowers in ! A difference of two volumes might be used to calculate the volumes of the sides of the vase, but what about the base ? Perhaps he just meant vase-shaped solid.

So lets assume we just have calculate the all solid volume since the main point of the question was to ask which integral formula to use.

Short answer: We use the concept of your first formula but adapted to be with respect to y.
V = pi∫[g(y)]^2 dy (see A2 below)

Imagine horizontal slices across this volume producing "elemental" circular discs centred on the y axis, each of thickness dy. (This reminds us that we are integrating with respect to y).
Although we can see that their areas are each pix^2, we want that in terms of y and since x^2 = y the volume integral is just V = pi∫y dy, with limits y = 0 to y = a^2
V = pi|(1/2)y^2| {y: 0 to a^2} = (pi/2)a^4

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DISCS (RINGS) WASHERS AND SHELLS: The differences between these formulae.

DISC methods: with respect to x or with respect to y
A1) V = pi∫[f(x)]^2 dx.....................A2) V = pi∫[g(y)]^2 dy

Sometimes we need to calculate a volume as a difference between two volumes, based on the disc method, and the elemental volume (a large disc minus a small disc), is shaped like a washer, so although this is essentially the same method, it is referred to as the WASHER method.
Images are very helpful so I refer you to these two websites for the washer method.
http://www2.bc.cc.ca.us/resperic/Math6A/Lectures/ch6/2/washer.htm
See Example 2 in Paul's online notes
http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx

CYLINDRICAL SHELL methods: with respect to x or with respect to y
(volume element = circumference*height* thickness; e.g (2pix)*f(x)*dx)
B1) 2pi∫xf(x) dx .....................B2) 2pi∫yf(y) dy

For more images and an explanation of why and when we need the cylindrical shell method, see
http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-Volums-CylinShells_Stu%20.pdf

Regards - Ian

You use "∫ π [f(x)]^2 dx" if
You find the common cylinder that has a radius of f(x) and dx as the thickness, and if
The area under f(x) rotates about the x-axis.
(you may refer to http://img.sparknotes.com/figures/6/63c1992d178b65dd1202406ca81625b5/intapps2.gif )

You use "∫ 2πx f(x) dx" if
You find the common cylindrical shell that has a height of f(x), dx as the thickness, radius (along the x-axis) of x and thus circumference (or length if you stretch out the shell) of 2πx, and if
The area under f(x) rotates about the y-axis.
(http://img.sparknotes.com/figures/6/63c1992d178b65dd1202406ca81625b5/intapps1.gif )

For the vase, you use "∫ 2πx f(x) dx".

Hi,

One difference is that the first formula calculate the volume of the solid created by f(x) rotating about the x-axis and the latter formula calculate volume of the solid created by f(x) rotating about the y-axis

Another is the idea behind the formula of calculating volume. The first one is finding volume through dividing into many small disks while the second one is dividing into many small shells (rings ).

I think you should use the second formula

Hope that helps :>

For the first you draw strips of height y and thickness dx
then rotate this about the x-axis to get volume of a thin disk.
It has radius y and thickness dx with volume dV=πy^2 dx
and y=f(x) so you get V from ∫π[r(x)]^2dx
For the second draw the same strip as a above, then the area
of this strip is ydx . When rotated about the y-axis you get
a thin cylindrical tube, of radius x, length y and thickness dx.
Its volume dV=2πxydx=2πxf(x)dx and so
V=∫2πxf(x) dx.

In your problem you are rotating about the y-axis so use the second formula.
You should get (πa^4)/2

The first is correct, you want to take individual areas of disks, and then you multiply it the length, but that is if the length is true for everything, if not we integrate it over the length. The f(x) in your case is the x^2-g(x), where g(x) is just a bit smaller than f(x), because the vase has a thickness. However your instructor might want something similar to the equation. It wouldn't make sense however.