1000 Solved Problems in Modern Physics

56 1 Mathematical Physics

Fig. 1.15aArea bounded by ABGFA (see the text, Prob 1.53a )

Fig. 1.15bVolume of the
cylinder plus the cone
(See Prob 1.15b)

=π. 22

(

5 +

4

3

)

= 25

1

3

πunits

1.54 (a) Area=

∫ 2 π

0

ydx=

∫π

0

xsinxdx+

∫ 2 π

π

xsinxdx

=−xcosx+sinx|π 0 −xcosx+sinx|^2 ππ=π+ 3 π= 4 π

The area refers to the magnitude

(b) Volume,V= 2 π

∫

y^2 dx= 2 π

∫

x^2 sin^2 xdx

=

4 π^4

3

−

π^2

2