For a given density function fxy1 the total mass m in r is...

For a given density function f(x,y)=1, the total mass M in R is given by M=∬{R} f(x,y)dxdy and the center of gravity of mass in R has the coordinates x ̅ , y ̅ , where x ̅ =(1/M)∬{R} xf(x,y)dxdy , y ̅ = (1/M)∬{R} yf(x,y)dxdy. If R is the region between the circles x^2+y^2 = 1 and x^2+y^2=9 in the first quadrant, find M and x ̅.

Transcript

00:01 So in this question, we're told we have the following region.

00:06 Y.

00:07 We have x equals y squared here.

00:11 We have x plus y equals 2.

00:16 And we have x equals 2y.

00:20 And this is our region here.

00:23 This is r.

00:25 So where do these two, where do these things intersect? so we need to work out what this point and this point are.

00:32 So firstly for this point, we have x plus, so x equals 2, minus y and x equals 2y so 2 minus y equals 2 y so 3y equals 2 y equals 2 3y equals 2 thirds we also have that x equals 2 y so x equals 4 thirds now here we have y squared equals 2 minus y so that means that y squared plus y minus 2 equals 0 so y plus y plus 2 equals 0 so y plus 2y minus 1 equals 0.

01:23 So since we know that y is positive, we know that y is plus 1.

01:28 So this is 1.

01:30 And we also know that x plus y equals 2.

01:34 So x is also 1 here.

01:37 So sorry this is a slightly crowded figure now, but this has got all the information we need on it.

01:43 First of all, we need to get the total mass.

01:45 We're also told that the density is equal to x.

01:50 So the mass is going to be the integral, double integral.

01:56 We're going to integrate from y equals x over two to y equals the square root of x, x, d, y, d, x, and we're integrating from x equals zero to one.

02:13 But then we have to add on this bit here, which is the double integral, and we're integrating from y equals x over two to y equals two minus x, and from x equals one to x equals four thirds, x, d, y, y d x.

02:32 So this is going to be x root x so x to the three halves minus x squared over two integrated from nought to one with respect to x plus two minus x minus x over two this is going to be two minus three x over two so this is one to four thirds two x minus three x squared over two with respect to x.

03:07 So this is 2 5ths x to the 5 halves minus 1 6th x cubed from 0 to 1 plus x squared minus a half x cubed from 1 to 4 thirds so this is just going to be 2 5ths minus a 6th and this is going to be so 4 3rd squared is 16 9th and then we have to take away 1 and then here minus a half and then four thirds cubed.

03:46 So four cubed is 16 times four, which is 64, divided by 27, and then we have to add on a half.

03:56 So let's add all this stuff up.

03:59 Two -fifths minus a sixth, plus 16 over nine, minus one, minus a half, times 64 over 27.

04:15 Plus a half gives us 44 over 135.

04:24 So that's the total mass.

04:28 And now we want to get the x and y components at the center of gravity.

04:32 So the x component is going to be 1 over the mass, so 135 over 44...