What is the area enclosed by $r=theta$ for $theta in [0,pi]$ ?

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Answer 1 · 2016-05-07T11:54:13

The area = $pi^3/6=5.168$ , areal units.
Also, the length of the spiral $=[{(1/2)[pi sqrt (pi^2+1)+ln(pi+sqrt(pi^2+1)}-{0}]=6.110$ , nearly..

Area $=int(1/2)r^2 d theta=(1/2)int theta^2 d theta = (1/2)[theta^3]/3$ , between the limits, 0 and $pi$
$=pi^3/6=5.168$ areal units.

Added to the answer:
$r=theta$ . So, r' = 1
Length of the spiral = $int sqrt(r^2+(r')^2) d theta$

$=int sqrt(theta^2+1 )d theta$ , between the limits 0 and $pi$

$=.[(1/2)(theta sqrt( theta^2+1)+(1/2)ln(theta+sqrt(theta^2+1)]$ , between the limits 0 and $pi$

$=[{(1/2)[pi sqrt (pi^2+1)+ln(pi+sqrt(pi^2+1)}-{0}]=6.110$ , nearly..

What is the area enclosed by r=theta r=theta for theta in [0,pi]theta in [0,pi]?