54
1 Mathematical Physics
1.48 x
2/3
+ y
2/3
= a
2/3
(1)
The arc AB generates only one half of the surface.
S x
2
= 2π
b
a
y
1 +
dy
dx
2
1/2
dx
(2)
From (1) we find
dy
dx
= −
y
1/3
x 1/3 ; y =
a
2
3 − x
2
3
3/2
(3)
Substituting (3) in (2)
S x
2
= 2π
a
0
(a
2/3
− x
2/3 )
1 +
y
2/3
x 2/3
1/2
dx
= 2π
a
0
(a
2/3
− x
2/3 )
3/2
a
2/3
x 2/3
1/2
dx
= 2πa
1/3
a
0
(a
2/3
− x
2/3 )
3/2 x
−1/3 dx
=
6πa
2
5
∴ S x =
12πa
2
5
Fig. 1.13 Curve of
hypocycloid
x
2/3 + y
2/3 = a
2/3
1.49
a
0
√
a 2 −x 2
0
(x + y)dy dx =
a
0
√
a 2 −x 2
0
(x + y)dy
dx
=
a
0
x y +
y
2
2
dx
√
a 2 −x 2
0
=
a
0
x
a 2 − x 2 +
a
2
− x
2
2
dx
=
2a
3
3
1.50 Area to be calculated is
A = ACFD = 2 × ABED