Solutions to the Homework Set on page 170
2(a). , xy dxdy = R (y x2 )j, dy = R 0 dy = 0:
p p
p
2(b). R 4 R,36 sin x cos y dxdy = R 4 (, cos y cos x)j,3 6 dy = int 4 ( , ) cos y dy = ( , ):
R1R1
0
1
0
1
1
2
0
1
0
1
2(c). R R x x y dydx = R (x y2 )jx dx = int x5 dy = :
2(d). R, R y x y , xy dxdy = R, (y x4 , y x2 )jy dy = int,
2
0
2
1
2
0
3
0
3
0
3
2
3
1
p1
p1
3
4
2
2
3
2
1
2
16
3
2
0 2
0
2
1
2
3
2
0
0
2
0
2
y5
1 4
, y5 dy = ,
2
21
:
8
2(e). R R ,x2 y dydx = R ( y2 )j ,x2 dx = R 1 , x dx = :
2(f). R,aa R,x, x y dydx = 0 because the inner integral is of the form Rbb f (y) dy which is zero
1
0
1
0
0
1
1+
(why?).
2
1
0
1
2
0
1
3
2
2
3(a). R, R,, xy dydx = .
3(b). R R sinh x cosh y dydx = (cosh 1 , 1)(sinh 2 , sinh 1).
R
3(c). R R xyR sin(
x + y ) dydx: Note that y sin(x + y ) dy = , cos(x + y ) (let u =
y sin(x + y ) dy = , cos(x + y )j = , cos( + x ) + cos x . Finally,
R
R
Rx +
R y ). Hence,
x cos( +x ) dx+
x cos x dx = , sin( +x )j + sin x j
xy sin(x +y ) dydx = ,
= , sin 2 + sin .
3(d) R R x2xyy2 dydx = (13Ln 2 , 5Ln 5).
4(a) The equation of the plane that connects (1; 0; 0), (0; 1; 0) and (0; 0; 1) is x + y + z = 1. So
1
2
3
2
1
0
2
0
2
1
0
2
2
0
2
0
1
4
2
2
1
15
4
2
1
2
1
2
+
2
2
0
2
1
2
2
1
2
2
2
0
2
2
1
2
2
0
0
1
2
2
1
2
2
2
1
4
2
2
2
1
2
2
2
2
0
1
4
2
0
1
2
the volume of the tetrahedron can be found by computing the double integal
Z Z
D
1 , x , y dxdy
where D is the triangle D = f(x; y)j0 < x < 1 , y; 0 < y < 1g. Hence
Z 1 Z 1,y
V =
1 , x , y dxdy = 16 :
0
0
The integral is computed in Mathematica by entering
Integrate[1 - x - y, {y, 0, 1}, {x, 0, 1 - y}]
4(c) The equation of the plane that connects (a; 0; 0), (0; b; 0) and (0; 0; c) is xa + yb + zc = 1. So the
volume of the tetrahedron can be found by computing the double integal
Z Z
D
c(1
, xa , yb ) dxdy
where D is the triangle D = f(x; y)j0 < x < a(1 , yb ; 0 < y < bg. Hence
V
=
Z b Z a(1, yb
0
0
c(1
, xa , yb ) dxdy = abc
6 :
The integral is computed in Mathematica by entering
1
Integrate[c(1 - x/a - y/b), {y, 0, b}, {x, 0, a(1 - y/b)}]
5. The quantity R RD dxdy is the volume of the region bounded by D and the plane z = 1. This
value isRnumerically
equal to the area of D (why?).
R
(a) 0a 02 rdrd = r2 .
2
(b) The boundary of D is given by xa22 + yb2 = 1, So, let r and be related to x and y through
the relations
x = ar cos ; y = br sin :
The jacobian of this change of coordinates is abr. Therefore
area of D =
(c)
R1Rx
x2 dydx
0
= 16 .
Z 2 Z
0
1
0
abr drd
= ab:
6(a) jacobian = jdet 11 ,11 j = 2. The square in the xy-plane is mapped into a parallelogram
in the uv-plane. For example, the edge x = 1 is transformed to the edge u , v = 1 in the xy-plane.
6(b) jacobian = jdet 2u ,2v j = 2(u2 + v2 ).
v
u
u cos v , sinh u sin v
6(c) jacobian = jdet cosh
cosh u sin v sinh u cos v j = cosh u sinh u.
7(a) R R ,x x dydx = .
7(e) Try polar coordinates. R R r ddr = .
8(b) The boundary of the region suggests changing coordinates to
r
p
3
1
0
1
0
1
6
1
0
3
0
4
x=
2 r cos t; y = 3r sin t:
Compute the jacobian J and proceed to evaluate the integral as
Z
1
0
Z 2
0
3 r2 cos2 tJ dtdr:
2
2