What is the area inside the curve r = 2 sin Θ and between the rays Θ = π/12 and Θ = 3π/4 ?
a. 0.0788
b. 0.2736
c. 0.6024
d. 1.0472
e. 1.5590
f. 2.0708
g. 2.5156
h. 2.8444
i. 3.0392
j. 3.1180
or is it none of these ?
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Answers & Comments
Verified answer
We want to integrate 2 sin O, so here's what we do...
the limits on our integral are pi/12 and 3pi/4... other than that, this is just a regular old integral... 'cept in polar coordinates.
integral(2 sin 0) = 2 integral(sin 0) = 2 -cos 0 from pi/12 to 3pi/4...
-2(cos 3pi/4 - cos pi/12) = -2(-1/sqrt(2) - cos(pi/12))
= 2/sqrt(2) + 2cos(pi/12) = sqrt(2) + 2cos(pi/12) = 3.346
permit x = theta (because of the fact i do no longer would desire to make a theta image) If tan x = sin x / cos x, then merely divide the two factors by making use of sin x. then you definately get one million / cos x = 2 while you're making one edge merely cos x, then you definately get cos x = one million/2 so as meaning everywhere you have x over the hypotenuse equals one million/2, you have an answer. that still skill you go with a favorable x value. so which you have your solutions at 60 and 3 hundred ranges. additionally, bear in mind which you will bypass around the circle greater effective than 360 ranges (it might in reality propose you already went around the circle. So the genuine solutions are 60 + 360n and 3 hundred + 360n, the place n is any integer.
h. 2.8444