# Find the exact length of the polar curve. r = θ^{2}, 0 ≤ θ ≤ π?

**Solution: **

Given, r = θ^{2}, 0 ≤ θ ≤ π

We have to find the exact length of the polar curve.

The length of the polar curve is given by \(\int_{a}^{b}\sqrt{(r^{2}+(\frac{dr}{d\theta })^{2})d\theta }\)

Now, \(\frac{dr}{d\theta }= \frac{d(\theta )^{2}}{d\theta }=2\theta\)

Length of the polar curve = \(\int_{0}^{\frac{5\pi }{4}}\sqrt{((\theta ^{2})^{2}+(2\theta )^{2})d\theta }\)

= \(\int_{0}^{\pi }\sqrt{(\theta^{4}+4\theta^{2})d\theta }\)

Taking out common term,

= \(\int_{0}^{\pi}\sqrt{\theta ^{2}(\theta^{2}+4)d\theta }\)

= \(\int_{0}^{\pi}\theta \sqrt{(\theta^{2}+4)d\theta }\)

= \(\left [ \frac{(\theta ^{2}+4)^{\frac{3}{2}}}{3}\right ]_{0}^{\pi}\)

= \(\frac{1}{3}\left [ (\pi)^{2}+4)^{\frac{3}{2}}-(0+4)^{\frac{3}{2}} \right ]\)

= \(\frac{1}{3}\left [ (\pi^{2}+4)^{\frac{3}{2}}-(4)^{\frac{3}{2}} \right ]\)

= \(\frac{1}{3}\left [ (\pi^{2}+4)^{\frac{3}{2}}-8\right ]\)

Therefore, the exact length of the polar curve is \(\frac{1}{3}\left [ (\pi^{2}+4)^{\frac{3}{2}}-8\right]\).

## Find the exact length of the polar curve. r = θ^{2}, 0 ≤ θ ≤ π?

**Summary:**

The exact length of the polar curve r = θ^{2}, 0 ≤ θ ≤ π is \(\frac{1}{3}\left [ (\pi^{2}+4)^{\frac{3}{2}}-8\right ]\).

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