Question

Find the volume of the solid by rotating the region about y=4

y=secx and y=0 on the interval [0, π/3]

Answer 1

This one requires some work. Bear with me a little. Have you sketched all the lines and curves on the coordinate plane?

Answer 2

well I graphed 1/cos and 0

Answer 3

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.

Answer 4

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.

Answer 5

And a slab between y = 0 and y = 4 represents the whole thing. We need to find an area of the associated washer.

Answer 6

Ok I follow so far

Answer 7

We want to find the area of the green.

Answer 8

This gives us the area of our arbitrary "slab"... The area of the green is \[A(x)=\pi *4^2-\pi (4 - \sec x)^2\]

Answer 9

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.

Answer 10

\[V=\int\limits_{0}^{\pi/3}(-8\sec x+\sec^2 x)dx\]

This integral is mildly tough. Do you think you can handle it or do you want me to do it?

Answer 11

Forgot a pi.

Answer 12

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!! :)

Answer 13

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
\[-8\pi \int\limits_{0}^{\pi/3}\sec xdx+\int\limits_{0}^{\pi/3}\sec^2 xdx\]

Answer 14

The integral on the right is pretty easy. The one on the left is a bit tougher.

Answer 15

\[-8\pi \int\limits_{0}^{\pi/3}\sec xdx=-8\pi \int\limits_{0}^{\pi/3}[\sec x(\sec x+ \tan x)]/(\sec x + \tan x)dx\]

Answer 16

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\[-8\pi \int\limits_{u(0)}^{u(\pi/3)}(1/u)du\]

Answer 17

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 \[-8\pi \ln(2+\sqrt{3})\]

Answer 18

You still with me here?

Answer 19

Yes I'm still here :)

Answer 20

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.

Answer 21

Hold up. I realize I've made a mistake. I had my signs flipped the wrong way. So, take the volume as \[-(\sqrt{3}-8\pi \ln(2 + \sqrt{3}))=8\pi \ln(2 + \sqrt{3})-\sqrt{3}\]

Answer 22

so the answer is 8πln(2+√3)−√3?

Answer 23

Yep

Answer 24

Wait, I forgot the pi again.

Answer 25

Thank you so much for your help!!! I really appreciate it!

Answer 26

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: \[8\pi \ln(2+\sqrt{3})-\pi \sqrt{3}\]

Answer 27

THANK YOU!!! :)