Please help with this pesky integral.

[MAC-A-TAC]

Hello.
Please help simplify this integral. Here is the problem statement: Determine the coordinates of the centroid of a quarter of a circle (x[sup:3971tvmd]2[/sup:3971tvmd] + y[sup:3971tvmd]2[/sup:3971tvmd]) = R[sup:3971tvmd]2[/sup:3971tvmd] located in the first quadrant. Its area is ?R[sup:3971tvmd]2[/sup:3971tvmd] /4.

M[sub:3971tvmd]Y[/sub:3971tvmd] = ? x dA, where dA = y dx = ? R[sup:3971tvmd]2[/sup:3971tvmd]- x [sup:3971tvmd]2[/sup:3971tvmd] dx

Here is what I have worked out:
? dM[sub:3971tvmd]Y[/sub:3971tvmd] = ?[sub:3971tvmd]0[/sub:3971tvmd][sup:3971tvmd]R[/sup:3971tvmd] x ? R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd] dx

M[sub:3971tvmd]Y[/sub:3971tvmd] = 10x(R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd]) [sup:3971tvmd]3/2[/sup:3971tvmd] - 4(R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd]) [sup:3971tvmd]5/2[/sup:3971tvmd] /15


The next step in the answer key is:
M[sub:3971tvmd]Y[/sub:3971tvmd] = 1/3(R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd]) [sup:3971tvmd]3/2[/sup:3971tvmd]] [sub:3971tvmd]0[/sub:3971tvmd][sup:3971tvmd]R[/sup:3971tvmd]

The answer key gives M[sub:3971tvmd]Y[/sub:3971tvmd] = R[sup:3971tvmd]3[/sup:3971tvmd]/3 as the solution.

What do I need to do next to get from
M[sub:3971tvmd]Y[/sub:3971tvmd] = 10x(R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd]) [sup:3971tvmd]3/2[/sup:3971tvmd] - 4(R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd]) [sup:3971tvmd]5/2[/sup:3971tvmd] /15 to M[sub:3971tvmd]Y[/sub:3971tvmd] = 1/3(R[sup:3971tvmd]2[/sup:3971tvmd]- x[sup:3971tvmd]2[/sup:3971tvmd]) [sup:3971tvmd]3/2[/sup:3971tvmd]] [sub:3971tvmd]0[/sub:3971tvmd][sup:3971tvmd]R[/sup:3971tvmd]

Thank you for your help.

 

[Deleted member 4993]

Hello.
Please help simplify this integral. Here is the problem statement: Determine the coordinates of the centroid of a quarter of a circle (x[sup:2paz20hb]2[/sup:2paz20hb] + y[sup:2paz20hb]2[/sup:2paz20hb]) = R[sup:2paz20hb]2[/sup:2paz20hb] located in the first quadrant. Its area is ?R[sup:2paz20hb]2[/sup:2paz20hb] /4.

M[sub:2paz20hb]Y[/sub:2paz20hb] = ? x dA, where dA = y dx = ? R[sup:2paz20hb]2[/sup:2paz20hb]- x [sup:2paz20hb]2[/sup:2paz20hb] dx

Here is what I have worked out:
? dM[sub:2paz20hb]Y[/sub:2paz20hb] = ?[sub:2paz20hb]0[/sub:2paz20hb][sup:2paz20hb]R[/sup:2paz20hb] x ? R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb] dx

M[sub:2paz20hb]Y[/sub:2paz20hb] = 10x(R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb]) [sup:2paz20hb]3/2[/sup:2paz20hb] - 4(R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb]) [sup:2paz20hb]5/2[/sup:2paz20hb] /15<<< This is incorrect - how did you get this?

Substitute

u = R^2 - x^2

du = -2 x dx

Then continue (don't forget to change the limits of integration)....

The next step in the answer key is:
M[sub:2paz20hb]Y[/sub:2paz20hb] = 1/3(R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb]) [sup:2paz20hb]3/2[/sup:2paz20hb]] [sub:2paz20hb]0[/sub:2paz20hb][sup:2paz20hb]R[/sup:2paz20hb]

The answer key gives M[sub:2paz20hb]Y[/sub:2paz20hb] = R[sup:2paz20hb]3[/sup:2paz20hb]/3 as the solution.

What do I need to do next to get from
M[sub:2paz20hb]Y[/sub:2paz20hb] = 10x(R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb]) [sup:2paz20hb]3/2[/sup:2paz20hb] - 4(R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb]) [sup:2paz20hb]5/2[/sup:2paz20hb] /15 to M[sub:2paz20hb]Y[/sub:2paz20hb] = 1/3(R[sup:2paz20hb]2[/sup:2paz20hb]- x[sup:2paz20hb]2[/sup:2paz20hb]) [sup:2paz20hb]3/2[/sup:2paz20hb]] [sub:2paz20hb]0[/sub:2paz20hb][sup:2paz20hb]R[/sup:2paz20hb]

Thank you for your help.