Math 3435 Homework Set 11 Solutions 10 Points. x= 1,, is in the disk of radius 1 centered at origin


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1 Math 45 Homework et olutions Points. ( pts) The integral is, x + z y d = x + + z da where is = x + z 8 x + z = 4 o, is the disk of radius centered on the origin. onverting to polar coordinates then gives, π x + z y d = 8 6x + + 6z da = 8r + 6r dr dθ π ( 65 ) dθ ( 65 ) π 6 = =. (4 pts) A sketch of is to the right. Here are the parameterizations for each portion of the surface. r x, θ = xi + sinθ j + cosθk θ π, x z+ = cosθ + x= + z, yz, is in the disk of radius centered at origin (in polar of course) ( yz) x=,, is in the disk of radius centered at origin Now, do the integral for each of these surfaces. In this case we ll need to do the cross product stuff so let s get that taken care of first. = i rθ = cosθ j sinθk i j k rθ = = cosθk + sinθ j cosθ sinθ r r = cos θ + sin θ = x θ π + cosθ x d = x da = x dx dθ π + cosθ π π ( x ) dθ cos θ 8dθ = = =
2 Math 45 Homework et olutions Points π π x d = z + + da = r cosθ dr dθ = cos θ dθ = x d = + + da = 4 da = 4π The integral is here is just the area of the disk of radius.. o all together the integral is then, π 9 x d = + 4π = π 4. ( pts) First get the gradient, f( xyz,, ) = y x z + 9 f= 4 x,, 4x Notice that this is oriented in the positive y direction (because the y component is positive) so we ll need to use the negative of this. f = 4 x,, 4z The region comes from = x + z 9 x + z = 4 o, is a circle of radius centered at the origin. Now the dot product. F f = z, y, x 4 x,, 4z = 4xz y 4xy = y = ( x + z 9) Notice that I didn t bother with the f since they were just going to cancel when we go to do the integral. peaking of which, π π 8 64π F d = x + z 9 da = r r 9 dr dθ = dθ =. First, z = 4x 6y. The integral is then, Not Graded ( ) 8z x d = 8 4x 6y x da = x 48y da The region is to the right and the limits are, x y x+ omputing the integral gives, x+ 68 ( ) 8z x d = x 48y da = x 48y dy dx = 5 9 x + x 4 x dx = 9 5
3 Math 45 Homework et olutions Points 5. o, since we re using the surface from # let s get that info copied to here. r( x, θ) = xi + sinθ j + cosθk θ π, x + z = + cosθ x= + z, is the disk of radius centered at origin (in polar of course) x=, is the disk of radius centered at origin Now, go through each of the integrals. In this case we ll need r θ which from # is rθ = sinθ j + cosθk Note, that in this case this will always point outwards so we have the correct orientation. o the integral in this case becomes F d = x,, cosθ,sin θ, cosθ da = cos θda π + cosθ π dx d = cos θ θ = 4 cos θ cos θ dθ = 4π The gradient is f( xyz,, ) = x z f=,, This points in the positive x direction and so is pointing outward. The integral is now. F d = + z,, z,, da = + z da π π = r + r cosθ dr dθ = cosθ + dθ = π The gradient is f( xyz,, ) = x+ f=,, f=,, ince this is the back of the cylinder we need the orientation to be in the negative x direction to be outward. Note that we could just have easily done this directly using n = i. The integral is F d =,, z,, da = da = π As with # note that the integral is the area of a disk of radius o, the overall integral is F d = 4π + π + π = 6. o, in this case we ll use tokes Theorem in the following direction.
4 Math 45 Homework et olutions Points curl F d = F dr where is the boundary of the surface. o, this will be the circle of radius 8 that lies at x + y + z = 8 + z = z =± Note that we only need the positive since we are told that it is the upper half of the sphere. The parameterization of the curve is r t = 8 cos θ, 8 sin θ, r t = 8 sin θ, 8 cos θ, Now, do the dot product is, F r = y, y, x 8z 8 sin θ, 8 cos θ, ( y ) ( y) = 8 sinθ + 8 cosθ = 8 8 sin θ 8cosθsinθ The integral is then π curl F d = F dr = 8 8 sin θ 8cosθ sinθ dθ = 7. This time we ll use tokes Theorem in the following direction F dr = curl F d o, in this case we need the curl of the vector field. i j k curl F = F = = k k = k x y z y x z Now, there are a variety of surfaces we could use here but it seems like one of the easiest to use is y = 5 x + z and we ll need it to be oriented in the positive y direction (remember that the as we walk along the curve the surface must be on the left and our head will then point in the direction of the normal vector). The gradient is then, f xyz,, = y+ x + z 5 f= x,, z This gives the correct orientation and the region is x + y = 4. o, the disk of radius centered at the origin. The integral is then, F dr = curl F d =,, x,, z da = 6z da π π = r θ dr dθ ( θ) dθ = = Note that we used the polar conversions : 6 sin 6sin x= rcosθ, z = rsinθ. 8. We will be using the ivergence Theorem in the following direction.
5 Math 45 Homework et olutions Points F dr = E div FdV o we will need the divergence and E. div F = z + 8x z = 8x E is the portion of a sphere so we ll be doing this integral in spherical coordinates and the limits are, π π ϕ θ ρ The integral is then. π π F dr = 8x dv = 8ρ sin ϕcosθ dρdθ dϕ E π π π 8π 6sin ϕcosθ dθ dϕ 6sin d ϕ ϕ = = =
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