Math 53 Practice Midterm 2 A – Solutions
1. a) The area of the triangle is 2, so ¯y=1
2Z1
0Z2−2y
2y−2
y dxdy.
b) By symmetry ¯x= 0.
2. ρ=|x|=r|cos θ|. Using symmetry, I0=ZZD
r2ρ rdrdθ =
Z2π
0Z1
0
r2|rcos θ|rdrdθ = 4Zπ /2
0Z1
0
r4cos θdrdθ = 4Zπ /2
0
1
5cos θdθ =4
5.
3. x=x,y=x2, 0 ≤x≤1: so ZC
yx3dx +y2dy =Z1
0
x2x3dx + (x2)2(2xdx) = Z1
0
3x5dx =1
2.
4. a) Qx= 6x2+by2,Py=ax2+ 3y2.Qx=Pyprovided a= 6 and b= 3.
b) fx= 6x2y+y3+ 1 ⇒f= 2x3y+xy3+x+g(y). Therefore, fy= 2x3+ 3xy2+g′(y).
Comparing this with Q, we get 2x3+ 3xy2+g′(y) = 2x3+ 3xy2+ 2 so g′(y) = 2 and g= 2y+c. So
f= 2x3y+xy3+x+ 2y(+constant).
c) Cstarts at (1,0) and ends at (−eπ,0), so ZC
~
F·d~r =f(−eπ,0) −f(1,0) = −e−π−1.
5. a) ¯
¯
¯
¯
uxuy
vxvy¯
¯
¯
¯
=¯
¯
¯
¯
2x/y −x2/y2
y x ¯
¯
¯
¯
= 3x2/y. Therefore, dudv =|3x2/y|dxdy = 3|u|dxdy
and hence dx dy =1
3|u|du dv.
b) ZZR
dx dy =Z4
2Z5
1
1
3ududv =Z4
2
1
3ln 5 dv =2
3ln 5.
6. a) IC
Mdx =ZZR
−MydA. (Green’s theorem)
b) We want Msuch that −My= (x+y)2. We can use e.g. M=−1
3(x+y)3.
7. a) div ~
F= 2y, so IC
~
F·ˆ
nds =ZZR
2y dA =Z1
0Zx3
0
2y dydx =Z1
0
x6dx =1
7.
b) For the flux through C1,ˆ
n=−ˆ
implies ~
F·ˆ
n=−(1 + y2) = −1 where y= 0. The length
of C1is 1, so the total flux through C1is RC1(−1) ds =−1. The flux through C2is zero because
ˆ
n=ˆ
ıand ~
F⊥ˆ
ı.
c) ZC3
~
F·ˆ
nds =ZC1+C2+C3
~
F·ˆ
nds −ZC1
~
F·ˆ
nds −ZC2
~
F·ˆ
nds =1
7−(−1) −0 = 8
7.
8.
¡
¡
¡
¡
¡
1
1
Zπ/2
0Z1
0Z1
0
r2r dz dr dθ.
1
