Calculate the area of the right triangle using polar coordinates

Question

My solution was: $$\int_0^{π/3} \int_1^{\secθ} rdrdθ$$ However the correct solution is: $$\int_0^{π/3} \int_0^{\secθ} rdrdθ$$

I don't see the radius anywhere being zero, so I do not understand why the lower limit of the inner integral is zero. We also know that at $x = 1 \iff r\cosθ = 1 \underset{θ =0}{\iff} r = 1$

My question are:

1. Why is my integral wrong?
2. What would the area given by my integral look like?

Answer 1

Sine the region contains the origin $r$ should start from $0$. Your claim

$x = 1 \iff r\cosθ = 1 \underset{θ =0}{\iff} r = 1$

is not correct. Because $r \cos \theta =1$ represents only the line $x=1$. When you substitute $\theta = 0$ you are reffering to the point $(1,0)$ in cartisean coordinates.

Your integral represents the area of the given region when you intersect with it the unit circle.