Thomas Calculus 13th [Solutions]

Section 16.5 Surfaces and Area 1185
f[x_, y_]: {2x y, x 3y}
2 2
curve x 4y 4
ImplicitPlot[curve, [x, 3,3},{y, 2, 2}, AspectRatio Automatic, AxesLabel {x, y}];
ybonds Solve[curve, y]
{y1, y2] y/.bounds;
integrand: D[f(x,y][[2]],x]
Integrate[integrand, {x,
N[%]
D[f[x,y]],y]/Simplify
2, 2},{y, y1, y2}]
Bounds for y are determined differently in 45 and 46. In 46, note equation of the line from (0, 4) to (2, 0).
Clear[x, y, f]
2
f[x_, y_]: {x Exp[y], 4x Log[y]}
ybound 4 2x
Plot[{0, ybound}, {x, 0, 2.1}, AspectRatio Automatic, AxesLabel {x,y}];
integrand: D[f[x, y][[2]],x] D[f[x,y][[1]],y]//Simplify
Integrate[integrand, [x, 0, 2], {y, 0, ybound }]
N[%]
16.5 SURFACES AND AREA
1. In cylindrical coordinates, let
2
2 2 2
x r cos , y r sin , z x y r .
2
Then r( r, ) ( r cos ) i ( r sin ) j r k,
0 r 2, 0 2 .
2 2 2
2. In cylindrical coordinates, let x r cos , y r sin , z 9 x y 9 r .
2 2 2
Then r( r, ) ( r cos ) i ( r sin ) j 9 r k; z 0 9 r 0 r 9 3 r 3, 0 2 .
But 3 r 0 gives the same points as 0 r 3, so let 0 r 3.
x y
3. In cylindrical coordinates, let x r cos , y r sin , z z r .
2 2
Then r( r, ) ( r cos ) i ( r sin ) j r k . For 0 z 3, 0 r 3 0 r 6; to get only the first octant,
2
2
let 0 .
2
2 2
2 2
4. In cylindrical coordinates, let x r cos , y r sin , z 2 x y z 2 r.
Then r( r, ) ( r cos ) i ( r sin ) j 2 rk . For 2 z 4, 2 2r 4 1 r 2, and let 0 2 .
5. In cylindrical coordinates, let x r cos , y r sin ; since
2 2 2 2 2 2 2
x y r z 9 x y 9 r
2
2
z 9 r , z 0. Then r( r, ) ( r cos ) i ( r sin ) j 9 r k . Let 0 2 . For the domain of r:
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