Welcome, Guest. Please login or register.

August 20, 2019, 12:53:26 pm

Author Topic: Trig Functions Question  (Read 194 times)  Share 

0 Members and 1 Guest are viewing this topic.

georgebanis

  • Adventurer
  • *
  • Posts: 12
  • Respect: 0
Trig Functions Question
« on: February 12, 2019, 06:48:13 pm »
0
Hey guys, would you know how to do part B of the question below. I have the answer to part A (√3/2) but just need a hand with the second part where the answer is (b) π/24(4π +3√3).

The region R is bounded by the curve y = cos2x, the x-axis and the lines x = π/6 and x = −π/6. (a) Sketch R and then find its area. (b) Find the exact volume generated when the region R is rotated about the x-axis.

Thanks

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 305
  • \[F(s)=\int_0^\infty \!f(t)e^{-st}\,\text{d}t\]
  • Respect: +128
Re: Trig Functions Question
« Reply #1 on: February 12, 2019, 07:17:41 pm »
+2
Hey guys, would you know how to do part B of the question below. I have the answer to part A (√3/2) but just need a hand with the second part where the answer is (b) π/24(4π +3√3).

The region R is bounded by the curve y = cos2x, the x-axis and the lines x = π/6 and x = −π/6. (a) Sketch R and then find its area. (b) Find the exact volume generated when the region R is rotated about the x-axis.

Thanks

The volume of the specified region is given by \[V=\pi\int_{-\pi/6}^{\pi/6}\cos^2(2x)\;dx.\]Using double angle formulae, we can write this as \[V=\frac{\pi}{2}\int_{-\pi/6}^{\pi/6}\Big(1+\cos(4x)\Big)\;dx,\] which should be easy to evaluate.
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne


RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8449
  • "All models are wrong, but some are useful."
  • Respect: +2312
Re: Trig Functions Question
« Reply #2 on: February 12, 2019, 07:45:30 pm »
+2
The volume of the specified region is given by \[V=\pi\int_{-\pi/6}^{\pi/6}\cos^2(2x)\;dx.\]Using double angle formulae, we can write this as \[V=\frac{\pi}{2}\int_{-\pi/6}^{\pi/6}\Big(1+\cos(4x)\Big)\;dx,\] which should be easy to evaluate.
Note that double angles are not in the 2u syllabus.

Having said that.
Hey guys, would you know how to do part B of the question below. I have the answer to part A (√3/2) but just need a hand with the second part where the answer is (b) π/24(4π +3√3).

The region R is bounded by the curve y = cos2x, the x-axis and the lines x = π/6 and x = −π/6. (a) Sketch R and then find its area. (b) Find the exact volume generated when the region R is rotated about the x-axis.

Thanks
I can clearly see that Dan's method is correct, however this question was designed to be at the Extension 1 level. If it appeared in a 2U exam (and you weren't told to approximate it with say, Simpson's rule) then that is a mistake on their behalf. Otherwise, in the future, please post questions in the correct thread.