# Finding Areas

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Anyone got a clue how to do part b ? Normally these sorta finding area questions are fairly straight forward, but this involves a circle so I’m not too sure and the markscheme doesn’t really make sense to me.

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#2

The shaded area is the area of that part of the circle lying between the x-coordinates of Aand B *minus* the area under the parabola between those limits.

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#3

(Original post by

Anyone got a clue how to do part b ? Normally these sorta finding area questions are fairly straight forward, but this involves a circle so I’m not too sure and the markscheme doesn’t really make sense to me.

**Madman11**)Anyone got a clue how to do part b ? Normally these sorta finding area questions are fairly straight forward, but this involves a circle so I’m not too sure and the markscheme doesn’t really make sense to me.

but it can be easier if you parameterise the circle as

hence turning your integral into

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#4

Last edited by DFranklin; 3 weeks ago

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(Original post by

Area under the circle can be found via

but it can be easier if you parameterise the circle as

hence turning your integral into

**RDKGames**)Area under the circle can be found via

but it can be easier if you parameterise the circle as

hence turning your integral into

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#6

(Original post by

Thanks, but Ngl it’s a bit difficult reading that - don’t really understand what it says. Basically what I did was find the area under the parabola from 3 to -3 which gave me 16 and I was going to minus this from the area under the circle from the integral 3 to -3, but idk how to do that second part.

**Madman11**)Thanks, but Ngl it’s a bit difficult reading that - don’t really understand what it says. Basically what I did was find the area under the parabola from 3 to -3 which gave me 16 and I was going to minus this from the area under the circle from the integral 3 to -3, but idk how to do that second part.

imagine the sector OAB. For the sector, you can easily get the angle and hence get the area. Then there are two identical triangles either side which are easy to evaluate.

Equally the parametric integration as described by RDKGames shouldnt be hard, they're both just a couple of lines.

Last edited by mqb2766; 3 weeks ago

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How do u find the sector area ? It’s 1/2r^2theta, but Idk how to use it in this context ?

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#8

(Original post by

How do u find the sector area ? It’s 1/2r^2theta, but Idk how to use it in this context ?

**Madman11**)How do u find the sector area ? It’s 1/2r^2theta, but Idk how to use it in this context ?

Last edited by DFranklin; 3 weeks ago

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I think I’m just being an idiot but I genuinely don’t have a clue what you’re on about. Could you send me a pic of how you’d do the question - I feel I’d get a better idea of seeing a solution rather than words for this particular question. Thanks.

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#10

(Original post by

I think I’m just being an idiot but I genuinely don’t have a clue what you’re on about. Could you send me a pic of how you’d do the question - I feel I’d get a better idea of seeing a solution rather than words for this particular question. Thanks.

**Madman11**)I think I’m just being an idiot but I genuinely don’t have a clue what you’re on about. Could you send me a pic of how you’d do the question - I feel I’d get a better idea of seeing a solution rather than words for this particular question. Thanks.

See diagram below:

For the area of the sector AOB you need the radius (from O) and the angle, 2r.

Radius is given in the question.

To find r, look at the triangle AOD. It's one of the "standard" angles you should recognise from the values of the opposite and adjacent in that triangle.

**Note:**Angles must be in radians.

Hence 2r is? And hence area of sector is?

Then you just need to add the areas of the two triangles AEO and BOF, to get the area below the circle between A and B.

Last edited by ghostwalker; 3 weeks ago

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#11

Well, it does in principle require arcsin/cos to find the angle, it's just in this instance the angle is nice enough that you can do it by recognition instead.

(I realised it was a "nice angle" while thinking about how to talk them through drawing their own diagram but decided pointing it out without a diagram would probably only confuse).

(I realised it was a "nice angle" while thinking about how to talk them through drawing their own diagram but decided pointing it out without a diagram would probably only confuse).

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