Madman11
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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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DFranklin
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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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RDKGames
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(Original post by 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.
Area under the circle can be found via \displaystyle \int_{-3}^3 \sqrt{18 - x^2} \ dx

but it can be easier if you parameterise the circle as

x = 3\sqrt{2} \cos \theta
y = 3\sqrt{2} \sin \theta

hence turning your integral into

\displaystyle \int_{3\pi/4}^{\pi/4} 3\sqrt{2} \sin \theta \cdot (-3\sqrt{2}\sin \theta) \ d \theta
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DFranklin
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(Original post by RDKGames)
Area under the circle can be found via \displaystyle \int_{-3}^3 \sqrt{18 - x^2} \ dx
If you set X = (-3,0), then the area of this part of the circle can be found as is twice the area of the triangle OAX + the area of the sector OAB, which can all be found using only trig. (I suspect that integral is "too hard" for the OP).
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Madman11
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(Original post by RDKGames)
Area under the circle can be found via \displaystyle \int_{-3}^3 \sqrt{18 - x^2} \ dx

but it can be easier if you parameterise the circle as

x = 3\sqrt{2} \cos \theta
y = 3\sqrt{2} \sin \theta

hence turning your integral into

\displaystyle \int_{3\pi/4}^{\pi/4} 3\sqrt{2} \sin \theta \cdot (-3\sqrt{2}\sin \theta) \ d \theta
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.
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mqb2766
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(Original post by 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.
As DFranklin suggests, you can get the area by adding a sector area to 2 identical triangles.
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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Madman11
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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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DFranklin
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(Original post by Madman11)
How do u find the sector area ? It’s 1/2r^2theta, but Idk how to use it in this context ?
You'll need to use arcsin (or arccos) to work out theta given the information you have (you can easily calculate the radius of the circle, and you want twice the angle between A and x=0).
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Madman11
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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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ghostwalker
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(Original post by 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.
It's not quite as bad as DFranklin painted; doesn't require arcsin/arccos.

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.

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DFranklin
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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).
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