C4 Parametrics/Integration Question

Not entirely sure how to approach this question



However this is my attempt so far:



My initial idea was to multiply my answer by 4 to find the total area

I suppose finding the area of the square and taking that away could help, but you'd be left with the bits to the left and right of the square, which are in the circle but aren't in the square or the shaded region.

Your idea about multiplying 4 does seem right though. I'm not 100% sure, but the answer may involve finding two integrals (focus on one quadrant and multiply by 4 at the end) and taking one away from the other.

Not entirely sure how to approach this question



However this is my attempt so far:



My initial idea was to multiply my answer by 4 to find the total area

My thinking is you can turn the whole thing on it's side (treat y as x and vice versa) and do parametric integration from there. The problem is that it's not a function, and I can't remember how that works with regard to integrating for the area (whether it'll end up as 0 or not). Hopefully it gives you the magnitude of the area, in which case this should work, else I can't really think what the answer is!

As for treating y as x, you can see that I've done that in my working out! But are my limits definitely correct because I'm not entirely sure..

As for treating y as x, you can see that I've done that in my working out! But are my limits definitely correct because I'm not entirely sure..

I'm bad at reading other people's workings sorry :P

You've got the limits wrong - when you integrate parametrically you need to redo the limits, because the original limits would be $\displaystyle \int ^{\dfrac{\pi}{2}} _1$ and those are in terms of x, so you need to redefine them in terms of $\theta$ and then it should all work out.

I just did it and got 1.28, I don't know if that's right or not? I can post my working if it is and you want me to ~

The answer is 2.45673...

I can't quite see what your limits are, but they may be right. It's probably what you're integrating and why that's getting difficult.

If we look at the top right quadrant, and integrate between the bottom right point of the shaded area and the top left of that shaded area in that quadrant only, then that should give us the area between (1,root(3)/2), (0,2) and the x axis. If you take what you've just calculated from the integral between (1,0) and (0,2), then you get the area inside the curve that isn't shaded, which leaves you with one more step to find the area of the shaded area given that you have certain integrals and the area of the non-shaded region.

Sean, we strangely very close!

Basically I just shoved 2cosxcosx into my graphic calculator with the limits pi/6 and pi/2 and then multiplied that by 4 which gives me the correct answer however:

a) the answer is negative, which is confusing because I swear the correct order or limits is pi/6 and pi/2, no?

b) why don't we have to subtract the clear squares underneath the curves?

We are on the right lines though!

I believe this will give you the correct answer:

$4 \times \displaystyle\int^{\pi /6}_{\pi /2} x . \dfrac{dy}{d\theta}\ d\theta$

Try that.

Yeah I did that earlier (one sec, lemme copy and paste)

Basically I just shoved 2cosxcosx into my graphic calculator with the limits pi/6 and pi/2 and then multiplied that by 4 which gives me the correct answer however:

a) the answer is negative, which is confusing because I swear the correct order or limits is pi/6 and pi/2, no?

b) why don't we have to subtract the clear squares underneath the curves?

You have your limits the wrong way round then, mine are the opposite. So the integral I posted must be the correct one

I believe this will give you the correct answer:

$4 \times \displaystyle\int^{\pi /6}_{\pi /2} x . \dfrac{dy}{d\theta}\ d\theta$

Try that.

Why is it that way around, surely it should be pi/6 (indicates y=1) and then pi/2 (indicates x= 0)

Do you have any idea where I went wrong? thanks

I'm not to sure what you're trying to do? You seem to have an upper limit of a parameter value and a lower limit of a y value, which won't work