The second question! Area under parametric equations. Reddit Calculus 2

We will find the area of the region under the curve defined by parametric equations. This question is from Reddit / 69obx32zfj
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[@rufusjasko]

You know the point (k,8) is on the curve, so to find θ at x=k solve for y=8, which leads to (secθ)^3 = 8 which solves nicely.

[@kasai-_-2074]

Since the point P(K, 8) is given, instead of solving sqrt(3)/2*pi = 3*theta*sin(theta), one could simply set 8 = y =sec^3(theta).
This would then simplify to sec(theta) = 2 and you would only have to know where cos(theta) = 1/2, namely at pi/3.

[@dentonyoung4314]

My brain hurts now. That was complicated, though doable.

[@mikefochtman7164]

Just 'happens to be...' that the upper limit was pi/3. My old math teacher would say, "By inspection, we see that pi/3 is the solution to ...." lol That always used to bug me because once you see it, it's obvious. But if you don't see it, frustrating.

[@mil9102]

use the fact y = 8 at P to get θ

[@thesnackbandit]

Awesome.

[@Tom-zu2yc]

This was awesome. But the question seams to be to find the set of alfa, beta and lambda that works. So how do we know that's the only set? Or was the question to just find eny set that works?

[@Samir-zb3xk]

θ=0 to θ=π/3 are the only bounds on 0≤θ

[@user-rd6ib8sk6r]

This an A-Level question no? i remember doing it

[@akuntumbal1485]

Wow, that's genius

[@oraz.]

What a hard question wow

[@dougcree6486]

Calculus basics? Really?
The “teacher” that came up with this problem should buy a dictionary and/or a clue.