Math 225: Calculus III    Quiz 10    4/22&24/97

<<Calculus`Math225`
i={1,0,0};j={0,1,0};k={0,0,1};

Let Σ be the portion of the surface defined by x^2 + y + z == 4 that lies in the first octant. Express the surface area of Σ as an iterated integral in the variables y and z. (Do not evaluate the integral.)

Solution

Here's what the surface looks like:

xyz[a_,b_,c_] := Graphics3D[{Text["x",{a(1.05),0,0}],Line[{{0,0,0},{a,0,0}}],Text["y",{0,b(1.05),0}],Line[{{0,0,0},{0,b,0}}],Text["z",{0,0,c(1.05)}],Line[{{0,0,0},{0,0,c}}]}];

p1 = ParametricPlot3D[{Sqrt[4-y-t(4-y)],y,t(4-y)},
{t,0,1},{y,0,4}]

Show[xyz[3,5,5],p1,Axes->False,Boxed->False,ViewPoint->{3.126, 0.773, 1.040}]

The surface projects to the triangular region in the yz-plane defined by
    0 <= z <= 4-y
    0 <= y <= 4
We take the normal vector to this region to be p = i, and the area element is

g = Grad[x^2 + y + z]

a = Norm[g]/g.i

We must convert this to the y and z variables:

a /. {x^2->4 - y - z, x->Sqrt[4 - y - z]}

and the surface area of Σ is given by

Integrate[Sqrt[2 + 4(4 - y - z)]/(2Sqrt[4 - y - z]),
{y,0,4},{z,0,4-y}]

Let Σ be the portion of the plane 3x + 2y + z == 6 that lies in the first octant. Compute the surface integral x + 2y + z over Σ..

Solution

The surface is the level surface of f[x_,y_,z_] = 3x + 2y + z over the triangular region in the xy-plane:
    0 <= y <= (6-3x)/2
    0 <= x <= 2
This region is defined by the "octant" equations x==0, y==0, and the intersection of the given plane, 3x + 2y + z == 6, with the xy-plane, z == 0.
The normal vector to the region is p = k. The area element is given by

g = Grad[3x+2y+z]

Norm[g]/g.k

The integrand x+2y+z simplifies to x+2y+(6-3x-2y)==6-2x since the z==6-3x-2y on the surface. Thus the surface integral can be evaluated as

Integrate[(6-2x)Sqrt[14],{x,0,2},{y,0,(6-3x)/2}]

Show[{xyz[2.5,3.5,6.5],
Graphics3D[Polygon[{2i,3j,6k}]]}, Boxed->False,
ViewPoint->{3.165, -1.111, 0.447}]