Assignment 11 Solutions

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1 . Evaluate Math 9 Assignment olutions F n d, where F bxy,bx y,(x + y z and is the closed surface bounding the region consisting of the solid cylinder x + y a and z b. olution This is a problem for which the divergence theorem is ideally suited. alculating the divergence of F, we get F x, y, z bxy,bx y,(x + y z (x + y (b + z. Applying the divergence theorem we get F n d F dv (x + y (b + z dv π a b r (b + z r dz dr dθ b π a r dr dθ b ( a π πa b.. Find the flux of F x,y,z upward through the first-octant part of the cylindrical surface x + z a for y b. olution The surface is one of five surfaces that form the boundary of the solid region consisting of the part of the cylinder x +z a for y b that lies within the first-octant. The other four surfaces are plane surfaces: lies in the plane z, lies in the plane x, lies in the plane y, and lies in the plane y b. Orienting all surfaces so that the normal n points outwards we get n,,, n,,, n,,, n,,. If we let tot represent the total boundary of, i.e. tot, then we have F n tot d F n d + F n d + F n d + F n d + F n d tot F n d + ( z d + ( x d + ( y d + y d b d F n d F n d + b (area of On the other hand, using the divergence theorem, we get F n tot d F dv tot πa b F n d +. ( + y dv V + V ȳ,

2 Math 9 Assignment olutions where V πa b/ is the volume of, and ȳ b/ is the y-coordinate of the centroid of. The final result is F n d F n tot d πa b V + V ȳ πa b πa b. tot. Using the divergence theorem, evaluate (x + y d, where is the sphere x + y + z a. olution On we have n a x,y,z. We would like to choose a vector field F so that F n x + y. Observe that F ax,ay, will do. If is the interior of the sphere, then we have (x + y d F n d F dv adv a(volume of (a πa 8 πa.. alculate the flux of F x + y sin z,y + z sin x,z across the surface, where is the boundary of the solid bounded by the hemispheres z x y and z x y and the plane z. olution Using the divergence theorem we get: F n d F dv π π π π [ (x + y + dv (ρ sin ϕ + ρ sin ϕdρ dϕdθ ( sin ϕ + sin ϕ dϕdθ ( ( cos ϕ sin ϕ + 7 sin ϕ ( cosϕ + cos ϕ dϕdθ 7 ] π/ cosϕ dθ π [ ( + 7 ] dθ 9π.. alculate the flux of F xy,yz,zx across the surface, where is the boundary of the solid that lies between the cylinders x + y and x + y and between the planes z and z. olution This is a problem for which the divergence theorem is ideally suited. alculating the divergence of F,

3 Math 9 Assignment olutions we get Applying the divergence theorem we get F n d F dv F x, y, z xy,yz,zx x + y + z. π x + y + z dv (r + z r dz dr dθ π (r + r dr dθ 7π. 6. alculate the flux of F z x, y + tanz,x z + y across the surface, where is the surface z x y with normal pointing upwards. [Hint: Note that is not a closed surface. lose in the obvious way and then use the divergence theorem.] olution Given the complexity of the vector field F, trying to evaluate flux F n d directly is rather difficult and since the surface over which the integral is to be evaluated is not closed, the divergence theorem does not apply. However, consider the closed surface tot obtained by adding a flat bottom to the hemisphere. That is, let be the surface consisting of the portion of the xy-plane within the circle x + y, oriented with a downward pointing normal. Then let tot. Now, the integral we want to evaluate can be expressed as: F n d F n d F n d. tot The first integral on the right hand side is a surface integral over the closed surface tot and so the divergence theorem may be applied to this integral. Evaluating the divergence of F yields F x, y, z z x, y + tanz,x z + y x + y + z. Applying the divergence theorem we get F n d F dv tot π To evaluate the second integral, we parmeterize as follows. x r cosθ r : y r sin θ θ < π, which leads to z (x + y + z dv ρ sin ϕdρ dϕdθ π. R r cosθ,r sin θ, Rθ r sin θ,r cosθ, Rr cosθ,sin θ, N,, r On we have F N r(x z + y r sin θ. Evaluating the surface integral yields F n d π r sin θ dr dθ π ( cosθ dθ π π.

4 Math 9 Assignment olutions The final result is F n d tot F n d π F n d ( π π. 7. Evaluate F n d, where F xyz,x,e xy cosz and is the hemisphere z x y with upward pointing normal. olution We use tokes s theorem to convert this surface integral to a line integral around the boundary of the surface. The boundary of the surface is the circle x + y in the xy-plane. Parameterizing this circle in the obvious way, we get x cosθ, θ π { r cosθ,sin θ,, : y sin θ, which leads to d r sin θ,cos θ, dθ. z, Applying tokes theorem we get F n π d F d r cos θ dθ π ( + cosθ dθ (θ + sin θ π π. 8. Evaluate F d r, where F xy + y + e x,x + xy y + sin(e y,xz + sinh(z and is the curve formed by the intersection of the cone z x + y and the cylinder x + y 6 oriented in the clockwise direction when viewed from the origin. olution One can try to evaluate this line integral directly, but given the complexity of the vector field F, it may be easier to evaluate by using toke s theorem. alculating the curl of F, we get i j k F, z,y. xy + y + e x x + xy y + sin(e y xz + sinh(z The cylinder intersects the cone in a circle of radius in the plane z. We may take the surface for use in toke s theorem to be the inside of that circle. For we have R r cosθ,r sin θ, x r cosθ r Rr cosθ,sin θ, : y r sin θ θ < π, which leads to z Rθ r sin θ,r cos θ, N,,r

5 Math 9 Assignment olutions Applying toke s theorem we get F d r F n d π π F N dr dθ π [ r r r(r sin θ dr dθ sin θ dθ r ( dθ 6 π cosθ 8θ 6π ( 6 sin θ ] 9. Evaluate F d r, where F xy,x y,x + z and is the curve consisting of the line segments joining A(,, to B(,, to (,, back to A. olution One can try to evaluate this line integral directly, but this would require us to parameterize three separate line segments and splitting the integral into three. It may be easier to use toke s theorem. alculating the curl of F, we get F i j k,,. xy x y x + z The three line segments form a triangle. We may take the surface for use in toke s theorem to be the plane region inside of that triangle. A normal for the plane is given by N B BA i j k,, N n N,,. Applying toke s theorem we get F d r F n ( d d (area of. But the area of the triangle is half of the area of the parallelogram fromed by the vectors B and BA. Therfore the area is area of B BA,,. We finally get the result F d r (.. Evaluate F d r, where F (x + y,(x + y,yz and is the curve formed by the intersection of the cone z x + y and the cylinder (x +y oriented in the counterclockwise direction when viewed from high above the xy-plane.

6 Math 9 Assignment olutions 6 olution One can try to evaluate this line integral directly, but given the complexity of the vector field F, it may be easier to evaluate by using toke s theorem. alculating the curl of F, we get i j k F z,,. (x + y (x + y yz The cylinder intersects the cone in a closed curve that resembles a tilted ellipse. Given the geometry of the situation, cylindrical coordinates seem most appropriate. The equation for the cylinder is r cosθ, and the equation of the cone is z r. We may take for the surface is toke s theorem the portion of the cone that lies within the cylinder. For we have R r cos θ,r sin θ,r x r cosθ r cosθ : y r sin θ π θ < π Rr cosθ,sin θ,, which leads to Rθ r sin θ,r cos θ, z r N r cosθ, r sin θ,r Applying toke s theorem we get F d r F n d cos θ π/ 6 F N dr dθ ( r cos θ drdθ π/ cos 6 θ dθ [ + cosθ] dθ [ + cosθ + cos θ + cos θ ] dθ [ + cosθ + ] ( + cosθ + ( sin θ cosθ dθ [ + cosθ + ] cosθ sin θ cosθ dθ ] π/ [ θ + sin θ + 8 sin θ sin θ 6 [ ( π ] π.. Evaluate F d r, where F e x,x + sin(y,z and is the curve formed by the intersection of the cone z x + y and the cylinder x + (y oriented in the counterclockwise direction when viewed from high above the xy-plane. olution One can try to evaluate this line integral directly, but given the complexity of the vector field F, it may be easier to evaluate by using toke s theorem. alculating the curl of F, we get i j k F,,. e x x + sin(y z

7 Math 9 Assignment olutions 7 The cylinder intersects the cone in a closed curve that resembles a tilted ellipse. Given the geometry of the situation, cylindrical coordinates seem most appropriate. The equation for the cylinder is r sin θ, and the equation of the cone is z r. We may take for the surface in toke s theorem the portion of the cone that lies within the cylinder. For we have R r cosθ,r sin θ,r x r cos θ, r sin θ : y r sin θ, θ π, which leads to z r Applying toke s theorem we get F d r F n d π sin θ dθ π F N dr dθ [ θ ( cosθ dθ R r cosθ,sin θ, R θ r sin θ,r cosθ, N r cos θ, r sin θ,r π sin θ sin θ ] π π. r dr dθ. alculate the work done by a force field F x x + z,y y + y,z z + y when a particle moves under its influence around the edge of the part of the sphere x + y + z that lies in the first octant, in a counterclockwise direction when viewed form above. olution The work done by the force field F is given by F d r, where is the boundary of the portion of the sphere in the first octant. One can try to evaluate this line integral directly, but given the complexity of the vector field F and the curve, it may be easier to evaluate by using toke s theorem. alculating the curl of F, we get F i j k y,z,. x x + z y y + y z z + y The surface is the portion of the sphere in the first octant, which we parameterize as follows: R sin ϕcos θ,sin ϕsin θ,cosϕ x sin ϕcos θ, ϕ π/ R ϕ cosϕcos θ,cosϕsin θ, sin ϕ : y sin ϕsin θ, θ π/, which leads to z cosϕ, R θ sin ϕsin θ,sin ϕcos θ, N sin ϕcos θ,sin ϕsin θ,sin ϕcosϕ Applying toke s theorem we get F d r F n d F N dϕdθ sin ϕ(sin ϕsin θ cosθ + cosϕsin θ dθ dϕ sin ϕ(sin ϕ sin π/ θ cosϕcos θ dϕ 6 θ ( ( cos ϕ sin ϕ + sin ϕcosϕ dϕ. sin ϕ( sin ϕ + cosϕ dϕ