17.1

1.
x=cos(t)	y=sin(t)
dx=-sin(t)	dy=cos(t)
Int(Int(cos(t)*sin(t)*(-sin(t))+sin(t)*cos(t),r=0..1),t=0..2Pi)
=0
P=xy,	Q=y
Qx=0,	Py=x
Qx-Py=-x
Int(Int(-x,y=-sqrt(1-x)..sqrt(1-x)),x=-1..1)
=0
0=0
Therefore the Green theorem exist.

3.
P=y^2	Q=x^2
Qx=2x	Py=2y
Qx-Py=2x-2y
Int(Int(2x-2y,x=0..1),y=0..1)
=0

5.
P=x^2y	Q=0
Qx=0	Py=x^2
Qx-Py=-x^2
Int(Int(-x^2,y=-sqrt(1-x^2)..sqrt(1-x^2)),x=-1..1)
=-Pi/4

7.
Qx=2x	Py=0
Qx-Py=2x
Int(Int(2x,y=x^2..x),x=0..1)
=1/6

9.
Qx=e^(x-y)	Py=e^(x+y)
Qx-Py=e^(x-y)-e^(x+y)
(-1)Int(Int(e^(x-y)-e^(x+y),y=0..x),x=0..2)+Int(Int(e^(x-y)-e^(x+y),y=x-2..2),x=2..4)
=e^6/2-e^4/2-5e^2/2+5/2

13.
Qx=3	Py=1
Qx-Py=2
Closed ABCD=2*8*2/2=16
DA=Int(y,y=6..0)=-18
nonclosed ABCD=16-(-18)=34

17.2

1.
x=cos(t)	y=sin(t)	z=0
F=<2cos(t)sin(t), cos(t), sin(t)>
Fdr=(2cos(t)sin(t)(-sin(t))+cos^2(t))dt
Int(Fdr,t=0..2Pi)
=Pi
z=1-x^2-y^2
curl=<1,0,1-2x>
Int(Int(-2x+1-2x,y=-sqrt(1-x^2)..sqrt(1-x^2)),x=-1..1)
=Pi
Pi=Pi
Therefore the Stokes' Theorem exists.

3.
z=1
curl=<0,e^(y-z),-e^(y-z)>
=<0,e^(y-1),-e^(y-1)>
Int(-e^(y-1),y=0..1),x=0..1
=(e-1)*e^(-1)
Py=e^(y-z)	Qx=0
Qx-Py=-e^(y-z)=-e^(y-1)
Int(-e^(y-1),y=0..1),x=0..1
=(e-1)*e^(-1)
(e-1)*e^(-1)=(e-1)*e^(-1)

5.
curl=<-3ze^z^3,2ze^z^2+zsin(xz),2>
x=cos(t)	y=sin(t)	z=0
F=<1-sin(t), 1+cos(t), 1>
Fdr=((1-sin(t))*(-sin(t)+(1+cos(t))*cos(t))dt
Int(Fdr,t=0..2Pi)
=2Pi

9.
curl=<0,0,0>
Since F is conservative vector field, the answer is 0

11.
curl=<3,0,-5>
z=2
Int(Int(-5,y=-sqrt(3-x)..sqrt(3-x)),x=-3..3)
=-Pi/45

13.
plane: z=y
curl=<-1,-1,-1>
Fds=0dA
Int(Fds,dA)=0