54 1 Mathematical Physics

1.48x^2 /^3 +y^2 /^3 =a^2 /^3 (1)

The arcABgenerates only one half of the surface.

Sx

2

= 2 π

∫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

(^23)
−x

23 )^3 /^2

(3)

Substituting (3) in (2)

Sx

2

= 2 π

∫a

0

(a^2 /^3 −x^2 /^3 )

[

1 +

y^2 /^3

x^2 /^3

] 1 / 2

dx

= 2 π

∫a

0

(a^2 /^3 −x^2 /^3 )^3 /^2

(

a^2 /^3

x^2 /^3

) 1 / 2

dx

= 2 πa^1 /^3

∫a

0

(a^2 /^3 −x^2 /^3 )^3 /^2 x−^1 /^3 dx

=

6 πa^2

5

∴Sx=

12 πa^2

5

Fig. 1.13Curve of
hypocycloid
x^2 /^3 +y^2 /^3 =a^2 /^3

1.49

∫a

0

∫√a (^2) −x 2
0
(x+y)dydx=
∫a
0

[∫√

a^2 −x^2

0

(x+y)dy

]

dx

=

∫a

0

[(

xy+

y^2

2

)

dx

]√a (^2) −x 2
0

∫a
0

(

x

√

a^2 −x^2 +

a^2 −x^2

2

)

dx

=

2 a^3

3

1.50 Area to be calculated is

A=ACFD= 2 ×ABED