Solids of Revolution defined by inequalities

[Taniaz]

1. Find the centroids of the solids formed by rotating completely about the x-axis the plane regions defined by the following inequalities:
(a) y^2 < 9x, y>0, x<1
(b) xy<4, y>0, 1<x<2

2. I used the equation for solids of revolution:
Integral from a to b of (x[f(x)]^2.dx) / Integral from a to b [f(x)^2].dx3. I drew the graph and found the region enclosed as per their requirement. For a, I got the function in terms of y so y=3 sqrt (x) then I plugged it into the equation provided in 2. I took the bounds as x=0 and x=1 for the integration but I don't think this is the way to do it for inequalities.

 

[RUber]

What is it about the inequality that is throwing you off? Integration does not change between $\leq$ and $<$. Since the measure of the boundary point is zero, it neither adds nor subtracts from your volume.

 

[Taniaz]

This is what I did. I just took the function as 3 sqrt of x and integrated it from x=0 to x=1

y^2=9x
so y = 3 sqrt x
Is this correct?

 

[RUber]

Looks good to me. The most important thing for inequalities is to make sure you have the proper bounds. having y>0 and x<1 make nice bounds for this problem.

 

[RUber]

The problem asks for the centroid, so are you expected to give your answer as an ordered pair (x,y) or triple (x,y,z)? The x coordinate is the only non-trivial one.

 

[Taniaz]

I think since it's being rotated around the x-axis, they only require the x-coordinate?
Primarily I was confused with the region that was bound by y <3 sqt x. I first shaded the region as in the picture but then I was confused whether it was the region that intersects with the y-axis, above the curve of y=3 sqrt x or below it.

 

[Taniaz]

This is what I did for part b. More confused with this one region wise.

 

[RUber]

I think since it's being rotated around the x-axis, they only require the x-coordinate?
Primarily I was confused with the region that was bound by y <3 sqt x. I first shaded the region as in the picture but then I was confused whether it was the region that intersects with the y-axis, above the curve of y=3 sqrt x or below it.

Remember if it is y < f(x), you are looking at the area below the curve f(x). If you have x<f(y), you are looking at the region left of the curve f(y). You did it right.

This is what I did for part b. More confused with this one region wise.

This one looks like your region was defined properly as well. But you made an error in the math. log(2)/2 is not even in the region 1<x<2.
In one step you go from:
$\int_1^2 \frac{16}{x}dx = 4\log(x) |_1^2$ Do you see a problem with that?

 

[Taniaz]

Yes that's exactly why I thought I was looking at the wrong region. Let me fix it. Thank you!

 

[Taniaz]

So it's 2ln2 and not ln2/2 and that's about 1.39 which is between 1 and 2

 

[RUber]

Good work.
As you pointed out, the x coordinate of the centroid is the only one that is interesting. Clearly for a body of revolution about the x axis, the y and z coordinates of the centroid should be 0.

 

[Taniaz]

Thank you for your help!