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System Message

[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Tyrone (Customer)

[00:00:00] y=3x+1: x=0, x=2, y=0
rotating region about x-axis using disk method..final answer i got was 38pi

Julie N (Tutor)

[00:00:12] Welcome to Tutor.com ! How can I help you today?

Tyrone (Customer)

[00:00:22] Hello Julie

Julie N (Tutor)

[00:00:29] Hi

Tyrone (Customer)

[00:01:06] hope all is well this eveing...I
seem to be having slight
difficulty with a problem

Julie N (Tutor)

[00:01:35] alright, how would you like me to help?

[00:01:44] Can I see the integral you set up to evaluate this problem?

Tyrone (Customer)

[00:01:54] sure..

Julie N (Tutor)

[00:02:03] Wait a moment

[00:02:10] I got the same answer as you did

Tyrone (Customer)

[00:02:18] oh wonderful

Julie N (Tutor)

[00:02:24]

Tyrone (Customer)

[00:02:33] could you sketch the graph for
me just to confirm i did it
right?

Julie N (Tutor)

[00:02:40] Great work!

[00:03:41] Does that look familiar?

Tyrone (Customer)

[00:04:04] very!

Julie N (Tutor)

[00:04:13] That's just lovely

[00:04:29] What made you concerned with your answer?

Tyrone (Customer)

[00:04:56] just uncertainty..my biggest
problem is setting up the
problem

Julie N (Tutor)

[00:05:28] Would you like to show me the integral expression that you used or would you like to look at another problem so that you can set it up for me?

Tyrone (Customer)

[00:05:47] i would like to do another
problem if you dont mind

Julie N (Tutor)

[00:05:59] That would be great

Tyrone (Customer)

[00:06:05] thank you so much

Julie N (Tutor)

[00:06:26] Anytime!

Tyrone (Customer)

[00:06:41] its the same problem..except
now im rotating the region
along the line y=-1

[00:07:09] i can use any method to
solve..but i dont know how to
determine which method is
best or easiest

Julie N (Tutor)

[00:07:35] Have you learned both disc and shell method?

Tyrone (Customer)

[00:07:39] yes

Julie N (Tutor)

[00:07:45] Okay

Tyrone (Customer)

[00:07:56] im thinking washer

Julie N (Tutor)

[00:08:05] Perfect

Tyrone (Customer)

[00:08:10] yay!

Julie N (Tutor)

[00:08:34] Would you like to set up the integral for me?

Tyrone (Customer)

[00:08:43] yes absolutely

[00:08:53] im horrible with the pen
though

[00:08:55] lol

Julie N (Tutor)

[00:09:07] It's okay
I can follow

[00:10:37] To begin, I see you're already using 2pi...

[00:10:49] Make sure you're doing washers and not cylindrical shells.

Tyrone (Customer)

[00:11:01] ohhhh yes you're right!

[00:12:06] ?

Julie N (Tutor)

[00:12:15] Is that a 7 up top?

Tyrone (Customer)

[00:12:20] yes

Julie N (Tutor)

[00:12:36] Okay, once again, you're trying to use cylinidrical shells

[00:13:02] The difference between disks/ washers and shells is how your measuring the function

[00:13:12] and its relation to your axis of rotation.

[00:14:05] The value of the function is perpendicular to the axis of rotation.

Tyrone (Customer)

[00:14:23] using washer?

Julie N (Tutor)

[00:14:32] Yes

[00:14:54] The blue line represents the outer radius of the washer

Tyrone (Customer)

[00:15:02] right

Julie N (Tutor)

[00:15:16] In order to use the bounds from 0 to 7, we would have to be measuring from the y-axis.

[00:15:58] Rotating it about y = -1 would create that cylindrical shells since it's parallel to the axis of rotation.

[00:16:36] To do washers, we need to pick apart our cross section and identify the two radii

[00:17:47] Is this making sense/ ringing a bell?

Tyrone (Customer)

[00:17:55] kind of

[00:18:03] just need more practice i guess

Julie N (Tutor)

[00:18:28] The biggest thing with these problems is to identify which variable its easier to write the function in terms of

[00:18:44] right now, it's easiest to express y = 3x+1 in terms of x

Tyrone (Customer)

[00:18:53] right

Julie N (Tutor)

[00:19:11] so we'll have the bounds from 0 to 2, since those are the numbers x starts and stops at

Tyrone (Customer)

[00:19:30] yep makes sense

Julie N (Tutor)

[00:19:51] If we sketch in the "height" of our function, we see that when we rotate it we get the circles - which is why we went for washers

[00:20:07] So now, how can I write an expression for that outer radius

[00:20:13] (the blue strip)

Tyrone (Customer)

[00:21:17] hmmm...(3x+1)+2?

Julie N (Tutor)

[00:21:40] Close, I think my picture may have thrown you off

[00:22:49] So our outer radius should be (3x +1) +1

Tyrone (Customer)

[00:22:51] ohhhh nevermind i get it

Julie N (Tutor)

[00:22:55] or (3x + 2)

[00:22:56]

Tyrone (Customer)

[00:23:01] i miscounted

[00:23:03] lol

Julie N (Tutor)

[00:23:08] It's alright

[00:23:20] So, what is that inner radius?

Tyrone (Customer)

[00:23:36] so you can distrubute that 1?

Julie N (Tutor)

[00:23:52] For the inner or the outer radius?

Tyrone (Customer)

[00:23:57] the outer

[00:24:07] to get (3x+2)

Julie N (Tutor)

[00:24:09] yes, you can combine like terms

Tyrone (Customer)

[00:24:15] oh yeah duh

Julie N (Tutor)

[00:24:22] It's all good

Tyrone (Customer)

[00:24:24] i forget the basics

Julie N (Tutor)

[00:24:36] That's mostly the trouble with Calc

[00:24:48] The basics are always second-guessed

Tyrone (Customer)

[00:25:28] agreed!...the innner radius is
x?

Julie N (Tutor)

[00:25:45] Does that inner radius get larger as I move to the right?

[00:25:50] Or does it remain constant?

Tyrone (Customer)

[00:26:14] constant

Julie N (Tutor)

[00:26:26] Okay, so we can take the value as 1

[00:26:40] since over the entire interval, that inner radius is 1

Tyrone (Customer)

[00:26:54] oh yeah..now its clicking!

Julie N (Tutor)

[00:27:01]

Tyrone (Customer)

[00:27:14] may i set it up

Julie N (Tutor)

[00:27:20] i would love that!

[00:28:33] So far it looks wonderful!

Tyrone (Customer)

[00:28:50] thanks

Julie N (Tutor)

[00:28:52] Great!

[00:28:59] Just don't forget the dx at the end

[00:29:14] Perfect

Tyrone (Customer)

[00:29:25] lol..i think its a dx

Julie N (Tutor)

[00:29:33] I'll count it

Tyrone (Customer)

[00:29:36] yay!

[00:29:51] random question

[00:30:09] when using shell method..its
the opposite

Julie N (Tutor)

[00:30:26] What do you mean, the opposite...

Tyrone (Customer)

[00:30:58] well...i mean that radius is
perpendicular and height is
parellel

Julie N (Tutor)

[00:31:07] Yes

[00:31:38] and instead of adding circles (having pi times integral r^2)

Tyrone (Customer)

[00:31:52] ohhh thats right

Julie N (Tutor)

[00:31:56] we'd add up cross sections that are 2 pi * radius * heigh

[00:31:57] t

Tyrone (Customer)

[00:32:11] makes perfect sense

Julie N (Tutor)

[00:32:17] Awesome

Tyrone (Customer)

[00:32:36] would i get 38pi for the
answer?

Julie N (Tutor)

[00:33:05] Let me double check, but I want to say no...

Tyrone (Customer)

[00:33:10] ohhh ok

Julie N (Tutor)

[00:33:46] No you wouldn't get 38pi

[00:34:09] Even though it's the same region, the overall radius is stretched wider, then you are removing a cylinder from the center

Tyrone (Customer)

[00:34:15] oh..see my error

Julie N (Tutor)

[00:34:25] Simplifying the expression within the integral should clear it up a bit

[00:34:27] Yay!

Tyrone (Customer)

[00:35:02] ok i have a much better
understanding now..a little
more confident than before

Julie N (Tutor)

[00:35:14] That's great!

Tyrone (Customer)

[00:35:35] well thanks for all your help
Julie and have a great night

Julie N (Tutor)

[00:35:43] Anytime! You as well!

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