#Irina Mukhametzhanova, Section 24
#This homework is OK to post

#17.3 Homework
#Exercise 1
#The Divergence Theorem states that:
#Int(Int(F*dS, over S)) = Int(Int(Int(div(F)*dV)))
#So first, find the divergence of the vector field:
#div(F) = delta . F = d/dx(z) + d/dy(x) + d/dz(y) = 0
#Our integral, with the given bounds, turns out to be:
#Int(Int(Int(0,x=0..4),y=0..2),z=0..3) = 0
#ANSWER: The value of the flux is 0

#Exercise 3:
#The Divergence Theorem states that:
#Int(Int(F*dS, over S)) = Int(Int(Int(div(F)*dV)))
#So first, find the divergence of the vector field:
#div(F) = delta . F = d/dx(2*x) + d/dy(3*z) + d/dz(3*y) = 2
#Our integral, with the given bounds, turns out to be:
#Int(Int(Int(2,z=0..2),y=-sqrt(1-x^2)..sqrt(1-x^2)),x=-1..1)
#The volume integral of a constant is the volume of the region times that constant. The volume of our cylinder is:
#Pi*r^2*h = Pi*(1^2)*2 = 2*Pi
#So, our volume integral is equal to:
#2 * 2*Pi = 4*Pi
#ANSWER: The value of the flux is 4*Pi

#Exercise 5:
#The Divergence Theorem states that:
#Int(Int(F*dS, over S)) = Int(Int(Int(div(F)*dV)))
#So first, find the divergence of the vector field:
#div(F) = delta . F = d/dx(0) + d/dy(0) + d/dz(z^3/3) = z^2
#Using the given boundary, our new integral is:
#Int(Int(Int(z^2,z=-sqrt(1-x^2-y^2)..sqrt(1-x^2-y^2)),y=-sqrt(1-x^2)..sqrt(1-x^2)),x=-1..1)
#We can use Maple to calculate the integral:
int(int(int(z^2, z = -sqrt(-x^2 - y^2 + 1) .. sqrt(-x^2 - y^2 + 1)), y = -sqrt(-x^2 + 1) .. sqrt(-x^2 + 1)), x = -1 .. 1);
#Result: 4*Pi/15
#ANSWER: The value of the flux is 4*Pi/15

#Exercise 7:
#The Divergence Theorem states that:
#Int(Int(F*dS, over S)) = Int(Int(Int(div(F)*dV)))
#So first, find the divergence of the vector field:
#div(F) = delta . F = d/dx(x*y^2) + d/dy(y*z^2) + d/dz(z*x^2) = x^2 + y^2 + z^2
#Using the given boundary, our new integral is:
#Int(Int(Int(x^2 + y^2 + z^2,z=0..3),y=-sqrt(4-x^2)..sqrt(4-x^2)),x=-2..2)
#We can convert this to cylindrical coordinates:
#x = r*cos(theta), y=r*sin(theta), z=z, dV=r*dr*dtheta*dz
#So, our new integral is:
#Int(Int(Int((r^2+z^2)*r,r=0..2),theta=0..2*Pi),z=0..3) = Int(Int(Int(r^3 + z^2*r,r=0..2),theta=0..2*Pi),z=0..3) = 
# = Int(Int((r^4/4 + r^2*z^2/2,r=0..2),theta=0..2*Pi),z=0..3) = Int(Int(4 + 2*z^2,theta=0..2*Pi),z=0..3) = 
# = Int((4*theta + 2*z^2*theta, theta=0..2*Pi),z=0..3) = Int(8*Pi + 4*Pi*z^2,z=0..3) = 8*Pi*z + (4*Pi/3)*z^3,z=0..3 = 
# = 24*Pi + (4*9*Pi) = 24*Pi + 36*Pi = 60*Pi
#ANSWER: The value of the flux is 60*Pi

#Exercise 11:
#The Divergence Theorem states that:
#Int(Int(F*dS, over S)) = Int(Int(Int(div(F)*dV)))
#So first, find the divergence of the vector field:
#div(F) = delta . F = d/dx(x^3) + d/dy(0) + d/dz(z^3) = 3*x^2 + 3*z^2
#Using the given boundary, our new integral is:
#Int(Int(Int(3*x^2 + 3*z^2,z=0..sqrt(4-x^2-y^2)),y=0..sqrt(4-x^2)),x=0..2)
#We can convert this to spherical coordinates:
#x = r*sin(phi)*cos(theta), y=r*sin(phi)*sin(theta), z=r*cos(phi), dV=r^2*sin(phi)*dr*dtheta*dphi,
#r^2 = x^2 + y^2 + z^2
#So, our new integral is:
#Int(Int(Int((3*r^2*sin(phi)^2*cos(theta)^2 + 3*r^2*cos(phi)^2)*r^2*sin(phi),r=0..2),theta=0..Pi/2),phi=0..Pi/2) 
#We can use Maple to calculate the integral:
int(int(int((3*r^2*sin(phi)^2*cos(theta)^2 + 3*r^2*cos(phi)^2)*r^2*sin(phi), r = 0 .. 2), theta = 0 .. Pi/2), phi = 0 .. Pi/2);
#Result: 32*Pi/5
#ANSWER: The value of the flux is 32*Pi/5

#Exercise 15:
#The Divergence Theorem states that:
#Int(Int(F*dS, over S)) = Int(Int(Int(div(F)*dV)))
#So first, find the divergence of the vector field:
#div(F) = delta . F = d/dx(x+y) + d/dy(z) + d/dz(z-x) = 1 + 1 = 2
#Using the given boundary, our new integral is:
#Int(Int(Int(2,z=0..9-x^2-y^2),y=-sqrt(9-x^2)..sqrt(9-x^2)),x=-3..3) = 
# = Int(Int(18 - 2*x^2 - 2*y^2,y=-sqrt(9-x^2)..sqrt(9-x^2)),x=-3..3)
#Convert to polar coordinates:
#x = r*cos(theta), y = r*sin(theta), r^2 = x^2+y^2, dA = r*dr*dtheta
#Int(Int((18-2*r^2)*r,r=0..3),theta=0..2*Pi) = Int(Int(18*r - 2*r^3,r=0..3),theta=0..2*Pi) = 
# = Int((9*r^2 - (1/2)*r^4,r=0..3),theta=0..2*Pi) = Int(81 - 81/2,theta=0..2*Pi) = Int(81/2,theta=0..2*Pi) = 81*Pi
#ANSWER: The value of the flux is 81*Pi