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I am having trouble with this area problem, I need to find the area of the region bounded by the curve:

$$r=\theta^2;0\leq\theta\leq \pi/4$$.

The area of the region is given by the equation $A=\displaystyle \int_a^b \frac{1}{2}r^2 d\theta$. Now I'm stuck finding the a and b i.e the limits of integration. Looking at the graph is clear the $a=0$, What I tried to do to find b: $$\theta^2=arctan(\frac{\pi}{4})$$ $$\therefore \theta=\sqrt{arctan(\frac{\pi}{4})}\approx0.82$$ How do I figure out the limits of integration for areas in polar coordinates ?

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$r$ at the boundary of the bounded region depends on $\theta$.

$\theta$ takes value from $0$ to $\frac{\pi}{4}.$

Hence the area can be found with the following expression:

$$\int_0^\frac\pi4 \int_0^{\theta^2} r\,dr\,d\theta$$

Remark about your expression: It seems to suggest that the final answer should still contain $r$ which should not be the case.

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  • $\begingroup$ Thanks for answering, I see the answer it's a double integral, problem is I am taking Calculus II and we have not seen a double integral in the whole semester, is there a way to solve the problem without double integration ? $\endgroup$ – Andres Romero Nov 4 '17 at 2:32
  • $\begingroup$ ..., well, if you evaluate the inside integral first, it becomes a single integral. $\endgroup$ – Siong Thye Goh Nov 4 '17 at 2:36

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