phy1020.DVI

where in the last step we movedR^2 =2outside the integral because it’s a constant. Now evaluate theintegral:
Z2

0

ZR

0

rdrdD

R^2

2

Z2

0

d (4.79)

D

R^2

2



ˇ

ˇ

ˇ

ˇ

ˇ

2

0

(4.80)

D

R^2

2

.20/ (4.81)

DR^2 (4.82)

And again we have derived the classical formula for the area of a circle.

Area of a Trapezoid

Suppose we have a trapezoid consisting of a side along thexaxis, two parallel vertical sides atxD 0 and
xDh, and a slanted top side that is a straight line. Let the vertical side atxD 0 have lengtha, and the
vertical side atxDhhave lengthb. Then the classical formula for the area of a trapezoid is the mean of the
lengths of the parallel sides times the distance between the parallel sides:

AD

aCb

2

h (4.83)

Let’s see if we can derive this formula from integral calculus. The slanted top side of the trapezoid passes
through the points.0;a/and.h;b/. It therefore has equation

.ya/D

ba

h 0

.x0/ (4.84)

or

yD

ba

h

xCa (4.85)

Using integral calculus, the area of the trapezoid is then the area under this line:
Zh

0



ba

h

xCa



dxD

ba

h

Zh

0

xdxCa

Zh

0

dx (4.86)

D



ba

h

x^2

2

Cax

ˇˇ

ˇ

ˇ

ˇ

h

0

(4.87)

D



ba

h

h^2

2

Cah







ba

h

02

2

Ca.0/



(4.88)

Dh



ba

2

Ca



(4.89)

Dh



ba

2

C

2a

2



(4.90)

D

aCb

2

h (4.91)