#Irina Mukhametzhanova, Section 24 #This homework is OK to post #16.4 Homework #Exercise 7: #The tangential vectors are partial derivatives of the surface with respect to a certain #variable. So, T_u is a partial derivative of G with respect to u, ans T_v is a #partial derivative of G with respect to v: #T_u = <2, 1, 3> #T_v = <1, -4, 0> #To find the normal vector N(u,v), we need to find a vector that is perpendicular to both tangential vectors. So, we take the cross product of T_u and T_v: #T_u x T_v = |i j k| # |2 1 3| # |1 -4 0| = # = i*(0+12) - j*(0-3) + k*(-8-1) = <12,3,-9> #To find the equation of a plane, you need one of its points. So, plug in u=1 and #v=4 into G(u,v) to find a point: #G(1,4) = (2+4,1-16,3) = (6,-15,3) #If you have the plane's normal vector and a point (x0,y0,z0) on the plane, #the plane's equation is: #n1*(x-x0) + n2*(y-y0) + n3*(z-z0) = 0 #Plug in <12,3,-9> and the point (6,-15,3): #12*(x-6) + 3*(y+15) - 9*(z-3) = 0 #12*x - 72 + 3*y + 45 - 9*z + 27 = 0 #12*x + 3*y - 9*z = 0 #4*x + y - 3*z = 0 #ANSWER: T_u = <2,1,3>, T_v = <1,-4,0>, N(u,v) = <12,3,-9>, and the tangent plane's #equation is 4*x + y - 3*z = 0 #Exercise 13: #For the surface integral, we would first need to find its normal vector. To do so, #we find a vector that is perpendicular to both rates of change of the surface with #respect to u and v. #T_u = #T_v = <(-1)*u*sin(v), u*cos(v), 0> #T_u x T_v = |i j k| # |cos(v) sin(v) 1| # |(-1)*u*sin(v) u*cos(v) 0| = # = i*(0 - u*cos(v)) - j*(0 + u*sin(v)) + k*(u*cos(v)^2 + u*sin(v)^2) = # = <(-1)*u*cos(v), (-1)*u*sin(v), u> #The dS in the integral is equal to ||N||. So, #dS = sqrt(((-1)*u*cos(v))^2 + ((-1)*u*sin(v))^2 + u^2) = sqrt(u^2 + u^2) = sqrt(2)*u #So, our new integral is: #Int(f*dS, over S) = int(z*(x^2+y^2)*sqrt(2)*dudv,over S) #We would need to convert the f(x,y,z) function into f(u,v). Use the paratemric #representation of the surface to do this: #f(x,y,z) = f(u*cos(v), u*sin(v), u) = u*(u^2*(cos(v)^2+sin(v)^2)) = u^3 #Finally, use the result and the u and v bounds to make the final integral: #int(int(sqrt(2)*u^4,u=0..1),v=0..1) = int(((sqrt(2)/5)*u^5,u=0..1),v=0..1) = # = int(sqrt(2)/5,v=0..1) = sqrt(2)*v/5,v=0..1 = sqrt(2)/5 #ANSWER: The value of the surface integral is sqrt(2)/5 #Exercise 15: #We can parametrize the surface given in terms of x and z: #G(x,z) = #For the surface integral, we would first need to find its normal vector. To do so, #we find a vector that is perpendicular to both rates of change of the surface with #respect to x and z. #T_x = <1, 0, 1> #T_z = <0, -2*z, 1> #T_x x T_z = |i j k| # |1 0 0| # |0 -2*z 1| = i*(0+0) - j*(1-0) + k*(-2*z-0) = <0, -1, -2*z> #The dS in the integral is equal to ||N||. So, #dS = sqrt(0^2 + (-1)^2 + (-2*z)^2) = sqrt(4*z^2 + 1) #So, our new integral is: #Int(f*dS, over S) = int(z*sqrt(4*z^2+1),over S) #With our x and z bounds, our final integral is: #int(int(z*sqrt(4*z^2+1),x=0..3),z=0..3) = int(z*x*sqrt(4*z^2+1),x=0..3,z=0..3) = # = int(3*z*sqrt(4*z^2+1),z=0..3) -> Substitute u=4*z^2+1, du=8*z*dz -> # -> int((3/8)*sqrt(u),u=1..37) = (3/8)*(2/3)*u^(3/2),u=1..37 = # = (1/4)*u^(3/2),u=1..37 = (1/4)*37^(3/2) - 1/4 ≈ 56.02 #ANSWER: The value of the surface integral is about 56.02 #Exercise 19: #x^2 + y^2 = 4 represents a circle of radius 2, centered on the origin in the xy-plane. #So, we can parametrize the surface given in cylindrical coordinates: #G(theta,z) = <2*cos(theta),2*sin(theta),z> #For the surface integral, we would first need to find its normal vector. To do so, #we find a vector that is perpendicular to both rates of change of the surface with #respect to theta and z. #T_theta = <(-2)*sin(theta), 2*cos(theta), 0> #T_z = <0, 0, 1> #T_x x T_z = |i j k| # |(-2)*sin(theta) 2*cos(theta) 0| # |0 0 1| = # = i*(2*cos(theta)-0) - j*((-2)*sin(theta)-0) + k*(0-0) = <2*cos(theta),2*sin(theta),0> #The dS in the integral is equal to ||N||. So, #dS = sqrt(2^2*(cos(theta)^2+sin(theta)^2)) = 2 #So, our new integral is: #Int(f*dS, over S) = int(2*exp(-z)),over S) #Because we include the full circle, our theta bounds are from 0 to 2*Pi. #Using our theta and z bounds, our final integral is: #int(int(2*exp(-z),theta=0..2*Pi),z=0..4) = int((2*theta*exp(-z),theta=0..2(Pi)),z=0..4) = # = int(4*Pi*exp(-z),z=0..4) = (-4)*Pi*exp(-z),z=0..4 = (-4)*Pi*exp(-4) + 4*Pi*exp(0) = # = 4*Pi*(1 - exp(-4)) #ANSWER: The value of the surface integral is 4*Pi*(1 - exp(-4)) #16.5 Homework #Exercise 5: #The formula for a surface integral oriented upwards would be: #Int(-P*dg/dx - Q*dg/dy + R, over S), where g(x,y) = z. #First, find g(x,y) by using the plane equation to solve for z: #3*x - 4*y + z = 1 #z = -3*x + 4*y + 1 #Next, find the partial derivatives g_x and g_y: #g_x = -3 #g_y = 4 #Using our vector field, P=y, Q=z, and R=x, plug all of the results into the integral: #Int(-y*(-3) - z*4 + x, over S) = Int(3*y - 4*z + x, over S) = # = Int(3*y - 4*(-3*x + 4*y + 1) + x, over S) = Int(3*y + 12*x - 16*y - 4 + x, over S) = # = Int(-13*y + 13*x - 4, over S) #Now we need to find what x and y values the projection is bounded by. It is given in the #question, so, 0<=x<=1 and 0<=y<=1. #Our final integral is: #int(int(-13*y + 13*x - 4,x=0..1),y=0..1) = int((-13*y*x + (13/2)*x^2 - 4*x,x=0..1),y=0..1) = # = int(-13*y + (13/2) - 4,y=0..1) = int(-13*y + 5/2,y=0..1) = (-13/2)*y^2 + (5/2)*y,y=0..1 = # = (-13/2) + (5/2) = -4 #ANSWER: The value of the surface integral is -4 #Exercise 7: #The formula for the line integral of a surface whose normal vector is pointing outward is: #Int(f(x,y,z)*||T_u x T_v||dudv,over S) #So first, we need to parametrize the surface. We can use spherical coordinates, since #the surface is a sphere of radius 3: #x = 3*cos(theta)*sin(phi), y = 3*sin(theta)*sin(phi), z=3*cos(phi), #where 0<=theta<=Pi/2 and 0<=phi<=Pi/2 #We first need to find the normal vector, or a cross product of the surface's rates of #change with respect to theta and phi: #T_theta = <(-3)*sin(theta)*sin(phi), 3*cos(theta)*sin(phi), 0> #T_phi = <3*cos(theta)*cos(phi), 3*sin(theta)*cos(phi), (-3)*sin(phi)> #T_phi x T_theta = |i j k | # |3*cos(theta)*cos(phi) 3*sin(theta)*cos(phi) (-3)*sin(phi)| # |(-3)*sin(theta)*sin(phi) 3*cos(theta)*sin(phi) 0 | = # = i*(9*cos(theta)*sin(phi)^2) - j*((-9)*sin(theta)*sin(phi)^2) + # + k*(9*sin(theta)^2*sin(phi)*cos(phi) + 9*cos(theta)^2*sin(phi)*cos(phi)) = # = i*(9*cos(theta)*sin(phi)^2) + j*(9*sin(theta)*sin(phi)^2) + k*(9*sin(phi)*cos(phi)) = # = <9*cos(theta)*sin(phi)^2, 9*sin(theta)*sin(phi)^2, 9*sin(phi)*cos(phi)> #Next, we need to represent the vector field as a vector of variables theta and phi: #F(theta,phi) = <0,3,3*cos(theta)*sin(phi)> #Find the dot product of F(theta,phi) and the normal vector: #F . N = 0 + (27*sin(theta)*sin(phi)^2) + (27*cos(theta)*sin(phi)^2*cos(phi)) = # = 27*sin(theta)*sin(phi)^2 + 27*cos(theta)*sin(phi)^2*cos(phi) #So, we can use Maple to solve our final integral is: int(int(27*sin(theta)*sin(phi)^2 + 27*cos(theta)*sin(phi)^2*cos(phi),theta=0..Pi/2),phi=0..Pi/2); #Result: (27*Pi)/4 + 9, which also can be written as (27*Pi/12)*(3*Pi + 4) #ANSWER: The value of the surface integral is (27*Pi/12)*(3*Pi + 4) #Exercise 9: #The formula for a surface integral oriented upwards would be: #Int(-P*dg/dx - Q*dg/dy + R, over S), where g(x,y) = z. #Find, find the partial derivatives g_x and g_y, or z_x and z_y: #g_x = -2*x #g_y = -2*y #Using our vector field, P=z, Q=z, and R=x, plug all of the results into the integral: #Int(-z*(-2*x) - z*(-2*y) + x, over S) = Int(2*x*z + 2*y*z + x, over S) = # = Int(2*x*(9-x^2-y^2) + 2*y*(9-x^2-y^2) + x, over S) #We can rewrite this in polar coordinates, because 9-x^2-y^2 could be seen as a sphere with radius of 3: int(int(2*r^2*cos(theta)*(9-r^2) + 2*r^2*sin(theta)*(9-r^2) + r^2*cos(theta),r=0..3),theta=0..Pi/2); #Result: 693/5 #ANSWER: The value of the surface integral is 693/5 #Exercise 11: #The formula for a surface integral oriented upwards would be: #Int(-P*dg/dx - Q*dg/dy + R, over S), where g(x,y) = z. #Find, find the partial derivatives g_x and g_y, or (1-x-y)_x and (1-x-y)_y: #g_x = -1 #g_y = -1 #Using our vector field, P=y^2, Q=2, and R=-x, plug all of the results into the integral: #Int(-y^2*(-1) - 2*(-1) - x, over S) = Int(y^2 + 2 - x, over S) #Our x and y bounds turn out to be 0<=x<=1 and 0<=y<=1. So, our final integral is: #int(int(y^2 + 2 - x,x=0..1),y=0..1) = int((y^2*x + 2*x - x^2/2,x=0..1),y=0..1) = # = int(y^2 + 2 - 1/2,y=0..1) = y^3/3 + 2*y - y/2,y=0..1 = 1/3 + 2 - 1/2 = (2+12-3)/6 = 11/6 #ANSWER: The value of the surface integral is 11/6