Calculating Second Moment of Area about X and Y Axes - Example Problem

[Bradracer18]

Homework Statement

Ok, I have the normal Y(up) and x(to the right) axis. I've got the curve(starting at orgin) y^2=x. On the x axis, it goes to x =4 inches. So, that is the area I'm looking at(shaded).

Question states this:

Find the second moment of the shaded area with respect to
a) x - axis
b) y - axis

I'm not exactly sure how to do this...but I tried.

Oh, and I found the top of the curve, is at y=2.

Homework Equations

The Attempt at a Solution

Ix = int(y^2 * dA)

= int(from 0 to 2)(y^2 *(4-y^2) dy = 4.27in^4

Is this portion correct? I'm not sure

Thanks,
Brad

 

[Dick]

An integral dA is an integral over an area. So you will get a double integral over dx and dy. Look up an example.

 

[Bradracer18]

I have looked at examples...I'm fairly confident this is how they did it.

Where do you see that I went wrong? I think my equations are correct...

 

[Dick]

Yes, your basic equations are good. Sorry. I didn't realize you'd already done the x integration. But draw the figure and think about your limits of integration.

 

[Bradracer18]

I have the picture drawn, on my paper here.

I was thinking, that due to integrating in terms of y, I should integrate from zero to the height of the curve.

So, if x=4, then y = sqrt(4) = 2. This is why zero to 2 seemed logical to me.

 

[Dick]

Why not -2 to +2 in y? Does it say only y>=0? If you've included a picture, I can't see it.

 

[Bradracer18]

no...i didn't include a pic...I said, in the first post(confusing, I know)...that we are starting at (0,0)...like the axis is at the beginning of this curve(coming out of orgin)...I can post a pic, if you'd like...with paint.

 

[Dick]

Don't include a pic. It's probably not necessary. But can y be negative? I know x can't.

 

[Bradracer18]

Nope...both start at 0,0...then goes out to 4 inches on the x axis...with that curve(y^2=x)...and need to find how tall it is(on the y-axis).

 

[Dick]

The curve y^2=x has both of the points (4,2) and (4,-2) on it. So the area enclosed between x=4 and x=0 has regions both above and below the x-axis. I'm asking if you are being specifically asked to not include the region below. If so then your answer is already correct.

 

[Bradracer18]

ok...Yes, I've only got a picture of quadrant 1 Guess that is how I should have put it, all along...

Ok...that is great...thank you!