# Find the centroid of the region bounded by the graph of y=cubeRoot{x^4 - 5}, the x axis, and the lines x=2 and x=3 about the x axis.Express your answer to 4 significant digits. Center of Mass of a...

Find the centroid of the region bounded by the graph of y=cubeRoot{x^4 - 5}, the x axis, and the lines x=2 and x=3 about the x axis.

Center of Mass of a Lamina and Centroid of a plane region.

Moments and Centroid of a plane region.

Theorem of Pappus for Volumes of Solids of Revolution.

grebjack | Certified Educator

I don't know what you're supposed to do with the notes at the bottom - maybe those were hints given with the problem?  I'm going to answer this question:  "Find the centroid of the solid created by revolving the region bounded by y = (x^4-5)^(1/3), y=0, x=2, and x=3 around the x-axis."

You should think of the centroid as the center of mass, or the balance point.  If you slice the solid straight through that point, the two pieces that result will have the same mass.  We can simplify this by thinking about where the middle is in the x-direction, in the y-direction, and in the z-direction.  Since your instructor was kind enough to make this a solid of revolution, it is symmetric around the x-axis.  So the middle y-value is 0, and the middle z-value is also 0.  The only value you need to find is the x-value that is the average of all the points that make up that region you were going to revolve around the x-axis.

If you sketch a graph, you'll notice that the region is a little taller on the x=3 side than it is on the x=2 side, so I expect the answer to be a bit closer to 3 than to 2.  If we had a finite number of points, we could just list all of their x-values, add those up, and divide by how many we had.  And there would be more x-values of 3 than of 2 because there are more points on the right boundary than on the left.

Since there are infinitely many points, we'll need to use a continuous analog.  To add up all the x-values of all the points, I'll multiply each x-value between 2 and 3 by the number of points with that x-value, which is the height of the region at that x-value.  Here is the integral that adds all of those up: (I'm using the capital S for an integral sign)

3
S x(x^4-5)^(1/3)dx
2

I want to divide by the total number of points I've just added up, which is the area of the region.  Here is the integral that finds the area of the region:

3
S (x^4-5)^(1/3)dx
2

You're asked to give a decimal approximation, so you don't have to worry about how to antidifferentiate this thing.  I used a TI calculator to estimate

fnInt(x(x^4-5)^(1/3),x,2,3)/fnInt((x^4-5)^(1/3),x,2,3) = 2.5514

which is a bit closer to 3 than to 2, as we predicted.  So now you know all three coordinates (x,y,z) of the centroid.