Need help on finding area (SOLVED)

[98.40_for_sure]

This is the very last question on the VCAA 08 Exam 2 Q5F

I've included in the attachment all the relevant information needed to find but i just can't seem to get it right. My method is to use an integral for the area on the negative side of the Re(z) axis and using 0.5pi(1)^2 on the right side. The answer is 2.51.

Any help would be greatly appreciated

PS This is what i used to try and work it out however it is wrong, can someone explain why it is wrong?

[Mao]

The integrand looks fine, the terminals should be the other way though, -rt[3]/2 to 0. That should give you a positive area.

Just checked it, your expression is fine, just switch the intervals.

[98.40_for_sure]

Ooh thanks mao i overlooked the terminals.

Anyway, when i changed it the other way around, the answer i get is 2.527 however... something's still wrong

EDIT: Oops i rechecked, the answer is actually 2.53

Cheers bro

[Mao]

When you have questions like this, it's often easier to transform it to something that's easier to find. If you break it down to three sections: the semicircle, a triangle and half a sector (curved bit up the top), you will find that the area of the semicircle is easy to find, area of the triangle is easy to find, and the remaining area can be given by due to the symmetry of the circle.

[98.40_for_sure]

When you have questions like this, it's often easier to transform it to something that's easier to find. If you break it down to three sections: the semicircle, a triangle and half a sector (curved bit up the top), you will find that the area of the semicircle is easy to find, area of the triangle is easy to find, and the remaining area can be given by due to the symmetry of the circle.

I couldn't see any way other than to split it into semicircle + other bit. Where's the triangle and half a sector?

[jasoN-]

I think this is what he meant, correct me if I'm wrong.

[98.40_for_sure]

Ohhhhhh yeah i get it now, thanks.

I just have a liking of finding area between two curves, much more simple