You will find it easiest if you write the integral in the form.
∫??∫??13 ,∫??∫??1r3drdθ ,that is, with θ on the "outside" integral. Looking at your diagram, the minimum θ value in the region of integration is =0θ=0, which occurs along the x axis. The maximum θ value occurs at the point (1,1)(1,1), giving =/4θ=π/4. So we have∫/40∫??13 .∫0π/4∫??1r3drdθ .For any fixed θ value, the values of r in the region go from the vertical line =1x=1 to the semicircle. At =1x=1 we have =secr=secθ as you have noted. The easiest way to get the maximum r value is to draw a line from the origin at angle θ until it hits the semicircle, and a line from there to (2,0)(2,0). Then you can see in the diagram a right-angled triangle (because the angle in a semicircle is a right angle) with hypotenuse 22, and so we get max=2cosrmax=2cosθ. So the integral is∫/40∫2cossec13 .∫0π/4∫secθ2cosθ1r3drdθ .For sketching the graph,=2−2‾‾‾‾‾‾‾√⇒2−2+2=0⇒(−1)2+2=1 ,y=2x−x2⇒x2−2x+y2=0⇒(x−1)2+y2=1 ,and not forgetting that ≥0y≥0.