Centroid

suspenc3

Ok, I think I've figured this out but the book gives me different answer for my x value:

[tex]y=4-x^2[/tex] [tex]y=0[/tex]

[tex]A= \int_0^2(4-x^2)dx[/tex] A=16/3

[tex]x bar= \frac{3}{16}\int_0^2 x(4-x^2)dx[/tex]

and this comes out to 3/4..The book says 0 for the X-value...where did I go wrong?

0rthodontist

Your region goes from -2 to 2 on the x-axis, not 0 to 2.

suspenc3

ohhhhhh ya..good poinT.

suspenc3

but wont that give me A=0?

Benny

Well...

[tex]
A = \int\limits_{ - 2}^2 {\left( {4 - x^2 } \right)dx} = \left[ {4x - \frac{{x^3 }}{3}} \right]_{ - 2}^2 = \left( {8 - \frac{8}{3}} \right) - \left( { - 8 + \frac{8}{3}} \right) \ne 0
[/tex]

BTW since you're finding the centroid of a 2D solid shouldn't you use a double integral? Or is there some sort of formula that you're already using.

pizzasky

Alternatively, you can deduce the x-coordinate of the centroid by symmetry.

suspenc3

oh..ok..No there is a formula..Find A..and then the X and Y coordinates

HallsofIvy

Yes, there is a formula: but that doesn't mean you can't use your intelligence!

The "centroid" of a figure is the geometric center. Since you figure is symmetric about the y-axis, obviously the centroid must be on that axis: the x coordinate of the centroid is 0.