MA2E01 Tutorial solutions #5
1. Use polar coordinates to evaluate the integral
1 √1−x2
cos(x2 + y 2 ) dy dx.
0
0
Switching to polar coordinates, we can write the given integral as
1
π/2 1
π/2 π/2
sin(r2 )
π sin 1
sin 1
2
r cos(r ) dr dθ =
dθ =
dθ =
.
2
2
4
0
0
0
0
r=0
2. Use polar coordinates to evaluate the integral
√ 2
2
0
8−y
y
dx dy
.
1 + x2 + y 2
Switching to polar coordinates, we can write the given integral as
π/4 √8
π/4 √
π/4
√8
r dr dθ
π
√
=
1 + r2
dθ =
2 dθ = .
2
2
r=0
1+r
0
0
0
0
1
y=x
x2 + y 2 = 1
2
x2 + y 2 = 8
1
√
8
Figure: The regions of integration for Problems 1 and 2.
3. The sphere of radius a around the origin is given by the parametric equation
r(θ, φ) = a sin φ cos θ, a sin φ sin θ, a cos φ ,
where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. Use this fact to compute its area.
To compute the area, we first need to find the normal vector rφ × rθ , namely
rφ = a cos φ cos θ, a cos φ sin θ, −a sin φ ,
rθ = −a sin φ sin θ, a sin φ cos θ, 0 ,
rφ × rθ = a2 sin2 φ cos θ, a2 sin2 φ sin θ, a2 sin φ cos φ = a sin φ · r.
Since r lies on the sphere of radius a, we have ||r|| = a and this implies
2π π
2π π
||rφ × rθ || dφ dθ =
a2 sin φ dφ dθ
Surface area =
0
0 2π 0
2π0
π
=
−a2 cos φ
dθ =
2a2 dθ = 4πa2 .
φ=0
0
0
4. A lamina with
√ density δ(x, y) = x + y is bounded by the x-axis, the line x = 1 and the
curve y = x. Find its mass and also its center of gravity.
The mass of the lamina is
1
M=
0
1
√
0
=
0
x
(x + y) dy dx =
x3/2 +
1
0
x
2x5/2
dx =
2
5
√x
y2
xy +
dx
2 y=0
1
13
x2
= .
+
4 0 20
Its center of gravity is the point (x0 , y0 ), where x0 is given by
√x
20 1 2
xy 2
x y+
x(x + y) dy dx =
dx
13 0
2 y=0
0
0
1
x2
190
20 2x7/2 x3
20 1
5/2
x +
=
dx =
+
=
13 0
2
13
7
6 0 273
20
x0 =
13
1
√
x
and y0 is similarly given by
√x
20 1 xy 2 y 3
y(x + y) dy dx =
dx
+
13 0
2
3 y=0
0
0
1
20 x3 2x5/2
6
20 1 x2 x3/2
+
dx =
+
= .
=
13 0
2
3
13 6
15 0 13
20
y0 =
13
1
√
x