# Area between 2 curves, Volume around X and Y, Centroid

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1. Apr 29, 2017

### DanRow93

g(x)= √(19x) = upper curve
f(x)= 0.2x^2 = lower curve

Firstly, I found the point of intersection, which would later give the upper values for x and y.
x=7.802
y=12.174

Then I found the area under g(x) and took away the area under f(x) to get the area between the curves.
31.67 units^2

This is where I will include the equations so it is easy to see if I have done it right.

VolX = ∫π[(g(x)^2)-(f(x)^2)] dx
(The outer line squared minus the inner line squared)
Putting in my values for g(x)^2 and f(x)^2
g(x)^2 = 19x
f(x)^2 = 0.04x^4
Between 0 and 7.802 this gave me a volume of 1090.133 units^3

VolY = ∫π[(f(y)^2)-(g(y)^2)] dy
This time the line that was the inner in relation to the x axis becomes the outer and vice versa.
f(y)^2 = 5y
g(y)^2 = (y^4)/361
Solving this integral between 0 and 12.174 gave me a volume of 698.59 units^3

Next, the task was to find the centroid of the area, specifically using the theorem of pappus. This task is given a distinction mark, however after looking at it it seems quite simple and I'm not sure if I am missing something.

VolX = Shaded area x 2π(ybar)

So (1090.133)/(31.67x2π) = ybar = 5.478

Then I haven't done xbar yet but I would think that I just do the same with VolY = Shaded area x 2π(xbar)

Is there anything that you can see that I might have missed that makes the last part more difficult?

Thanks!

2. May 1, 2017

### haruspex

I gather it is also bounded by x=0.
I don't understand why you are working with volumes to find the centroid of an area. In particular, how do you get π in there?