Polar curves/Area of rectangle

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Homework Statement


w4peKxQ.png


Homework Equations

The Attempt at a Solution



Part C is confusing me.

I got the height PQ to be 16/3root6

But I'm lost as to how to find the length SP. The mark scheme has the answer as 8, or (SP/2 = 4 therefore SP = 8) but I still can't figure it out, maybe it's 'cause I've been doing questions all day.

Can someone shed some light on this for me please.
 

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Try ##\theta=0## :rolleyes:
 
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BvU said:
Try ##\theta=0## :rolleyes:
<Content edited by mentor>
Thank you
 
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trew said:

Homework Statement


View attachment 225862

Homework Equations

The Attempt at a Solution



Part C is confusing me.

I got the height PQ to be 16/3root6

But I'm lost as to how to find the length SP. The mark scheme has the answer as 8, or (SP/2 = 4 therefore SP = 8) but I still can't figure it out, maybe it's 'cause I've been doing questions all day.

Can someone shed some light on this for me please.

Part (c) needs calculus, so should certainly not appear in a "pre-calculus" forum.
 
Ray Vickson said:
Part (c) needs calculus, so should certainly not appear in a "pre-calculus" forum.
Thread moved for this reason.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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