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

The area =

Also, the length of the spiral

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To find the area enclosed by the curve (r = \theta) for (\theta) in ([0, \pi]), you can integrate the polar function (r = \theta) over the given range of (\theta). The formula for the area enclosed by a polar curve is (A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta), where (r) is the polar function, and (\alpha) and (\beta) are the limits of integration for (\theta). In this case, (r = \theta) and (\alpha = 0, \beta = \pi). Thus, the area (A) can be calculated as follows:

[A = \frac{1}{2} \int_{0}^{\pi} (\theta)^2 d\theta]

Solving this integral yields:

[A = \frac{1}{2} \left[\frac{\theta^3}{3}\right]_0^\pi]

[A = \frac{1}{2} \left[\frac{\pi^3}{3} - \frac{0^3}{3}\right]]

[A = \frac{1}{2} \left[\frac{\pi^3}{3}\right]]

[A = \frac{\pi^3}{6}]

So, the area enclosed by the curve (r = \theta) for (\theta) in ([0, \pi]) is (\frac{\pi^3}{6}) square units.

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