anonymous
  • anonymous
Find the volume of the solid by rotating the region about y=4 y=secx and y=0 on the interval [0, π/3]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
This one requires some work. Bear with me a little. Have you sketched all the lines and curves on the coordinate plane?
anonymous
  • anonymous
well I graphed 1/cos and 0
anonymous
  • anonymous
Well, you should sketch the region between y = sec x and y = 0 and x = 0 and x = pi/3. If you put y = 4 on your graph. It should make sense. I'll draw something really quickly here.

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anonymous
  • anonymous
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anonymous
  • anonymous
From this, you want to realize that for an arbitrary "slab" you are going to have 4 - sec x in between y = 4 and the y = sec x that you DON'T want to count in your volume.
anonymous
  • anonymous
And a slab between y = 0 and y = 4 represents the whole thing. We need to find an area of the associated washer.
anonymous
  • anonymous
Ok I follow so far
anonymous
  • anonymous
We want to find the area of the green.
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anonymous
  • anonymous
This gives us the area of our arbitrary "slab"... The area of the green is \[A(x)=\pi *4^2-\pi (4 - \sec x)^2\]
anonymous
  • anonymous
If we simplify this, then A(x) = pi * (-8sec(x) + sec^2(x)). Then, there should be a volume formula in your textbook with \[V=\int\limits_{a}^{b}A(x)dx \] where A(x) is the area of our arbitrary "slab". We want from x = 0 to x = pi/3. Knowing A(x), it all comes down to an integral.
anonymous
  • anonymous
\[V=\int\limits_{0}^{\pi/3}(-8\sec x+\sec^2 x)dx\] This integral is mildly tough. Do you think you can handle it or do you want me to do it?
anonymous
  • anonymous
Forgot a pi.
anonymous
  • anonymous
I know how to integrate mostly, but I'm really bad at doing it with trig stuff. Would you mind showing me? Sorry for the trouble :( And thank you so much for your help!! :)
anonymous
  • anonymous
Okay. Here's the integral\[\int\limits_{0}^{\pi/3}\pi(-8\sec x+\sec^2 x)dx\] You can pull out some constants and split the integral to get \[-8\pi \int\limits_{0}^{\pi/3}\sec xdx+\int\limits_{0}^{\pi/3}\sec^2 xdx\]
anonymous
  • anonymous
The integral on the right is pretty easy. The one on the left is a bit tougher.
anonymous
  • anonymous
\[-8\pi \int\limits_{0}^{\pi/3}\sec xdx=-8\pi \int\limits_{0}^{\pi/3}[\sec x(\sec x+ \tan x)]/(\sec x + \tan x)dx\]
anonymous
  • anonymous
You see what I did there? I just multiplied by "1". Let u = secx + tanx, and du = secxtanx + sec^2(x), which is what is in the numerator. We get\[-8\pi \int\limits_{u(0)}^{u(\pi/3)}(1/u)du\]
anonymous
  • anonymous
Then, we get \[-8\pi \ln|\sec x + \tan x| + C\] as the antiderivative for the "left" part. Evaluated from x = 0 to x = pi/3, this quantity is \[-8\pi \ln(2+\sqrt{3})\]
anonymous
  • anonymous
You still with me here?
anonymous
  • anonymous
Yes I'm still here :)
anonymous
  • anonymous
All right, so we need to find the "right" part, i.e. \[\int\limits_{0}^{\pi/3}\sec^2 xdx\]. But this one is much easier because the antiderivative is tanx + C. Evaluated, we get \[\sqrt{3}\]. Add this to the other result and we get the "final" answer.
anonymous
  • anonymous
Hold up. I realize I've made a mistake. I had my signs flipped the wrong way. So, take the volume as \[-(\sqrt{3}-8\pi \ln(2 + \sqrt{3}))=8\pi \ln(2 + \sqrt{3})-\sqrt{3}\]
anonymous
  • anonymous
so the answer is 8πln(2+√3)−√3?
anonymous
  • anonymous
Yep
anonymous
  • anonymous
Wait, I forgot the pi again.
anonymous
  • anonymous
Thank you so much for your help!!! I really appreciate it!
anonymous
  • anonymous
There needs to be a pi in front of the square root of three after the minus sign. I sincerely apologize for the confusion, as I lose track having to write all these equations online: \[8\pi \ln(2+\sqrt{3})-\pi \sqrt{3}\]
anonymous
  • anonymous
THANK YOU!!! :)

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