Question

I have a sphere, x^2+y^2+z^2=25. I want to know the surface area above the cone described by z=sqrt(x^2+y^2). I have an idea of where to start, but...brain cramps. :P Notes to follow.

Answer 1

First, I put the equation into cylindrical coordinates, substituting in the value of z provided for the cone.

Answer 2

interesting writing. hahaha

Answer 3

:) I started out doing the Latex in Mathematica, but it was taking too long.

Answer 4

Now I take the derivatives of r(u,v) with respect to u and v:

Answer 5

Next the cross product. Maybe I should have done this in Mathematica! :)

Answer 6

Only, should I have subtracted the cone?

Answer 7

wait a minute

Your partial w.r.t u for the k component...

Answer 8

sin(u) becomes cos(u) when you take its partial wrt u, right?

Answer 9

You have 10sin(u) written there..

Answer 10

Oopsie. If only we had a machine that could...oh wait. :)

Answer 11

Aw man, I made even more biffs in in rv. Just a sec.

Answer 12

Speedy. (!)

Answer 13

I was just correcting your error! And I don't think you can combine \(\sf \hat k\)

Answer 14

Well maybe you can if you factor out \(\sf 5\cos(u)\) but idk

Answer 15

Wait a minute, shouldn't the first one be 5cos(u)cos(v) i ?

Answer 16

Oh yeah you're right,my bad.

Answer 17

I was focusedon the \(\sf \hat k\)

Answer 18

No worries, I'm just happy I can even half-way participate in the conversation.

Answer 19

\[\sf \vec r_u = [5\cos(u)\cos(v)]\hat i +[5\cos(u)\sin(v)]\hat j + [5\cos(u)\cos(v) +5\cos(u)\sin(v)]\hat k\]

Answer 20

Ok, I'm going to see if I get the same result on my partials, just a sec.

Answer 21

\[\sf \vec r(u,v) = [5\sin(u)\cos(v)]\hat i + [5\sin(u)\sin(v)]\hat j +[5\sin(u)\cos(v) +5\sin(u)\sin(v)]\hat k\]\[\sf \vec r_u = [5\cos(u)\cos(v)]\hat i +[5\cos(u)\sin(v)]\hat j + [5\cos(u)\cos(v) +5\cos(u)\sin(v)]\hat k\]\[\sf \vec r_v = [-5\sin(u)\sin(v)]\hat i + [5\sin(u)\sin(v)]\hat j +[5\sin(u)\sin(v)-5\sin(u)\cos(v)]\hat k\]

Answer 22

It kind of looked like you were trying to take second partials or something, deriving off of \(\sf \vec r_u\) or something.

Answer 23

This is not territory for the feint of heart (or ADD, in my case). :P

Answer 24

i'm not exactly sure how to go about solving this problem either, but i Can help correct your steps! (Learning this as well)

Answer 25

I appreciate the help, seriously. This semseter's been *brutal*.

Answer 26

And can you edit your question? Post what the exact question is/asking for?

Answer 27

Sure.

Answer 28

thanks

Answer 29

Ok, here's the setup for my cross product:

Answer 30

wait, can you post the actual question first, like word for word. I do not understand your question

Answer 31

This is the question:

Answer 32

Oh, okay. this makes more sense

Answer 33

http://www.math.uri.edu/~bkaskosz/flashmo/tools/graph3d/
http://www.math.uri.edu/~bkaskosz/flashmo/tools/graph3d/

Answer 34

Did you find the intersection first?

Answer 35

ohhh, makes alot more sense now

Answer 36

You have \[z=x^2 +y^2\] and \[z= \sqrt{25-x^2-y^2}\]

Answer 37

If you equate these two, you can find the intersection

Answer 38

Ok, but that's just for the bounds, right?

Answer 39

yep, bounds are important

Answer 40

Yes, of course. What I mean is, I still need to parameterize, differentiate and take the cross product, right? I'd be happy to find I'm overcomplicating it, though.

Answer 41

I don't see why you'd be taking a cross product right away... but ofc i'd have to solve the problem in order to see why.

Answer 42

What I'm looking for is the surface area of this:

Answer 43

The orange part.

Answer 44

Aw DERP. I waaay overcomplicated it!

Answer 45

I can set it up as if I wanted the surface of the entire sphere, take the cross product and only integrate from the intersection of the cone on the bottom to the top of the sphere.

Answer 46

Whoop, idk what happened there. Yeah you did overcomplicate it.

Answer 47

Ok, this shouldn't take much longer now. Wow, little misunderstandings can take up a TON of time. :P

Answer 48

ON the bright side, I guess I won't make that mistake again.

Answer 49

Ok, so here goes:

Answer 50

lol yea, you only need the intersection points :) then you can find the central angle

Answer 51

then you can find the surface area based off of the area of the circle, could potentially do it without calculus

Answer 52

Yeah, but when the final gets here in two week, I'd like to present it in the way the prof want to see it. :)

Answer 53

well... yea... you can do that using a surface integral

Answer 54

but you should get two identical answers

Answer 55

Working through it as we speak.

Answer 56

okee dokey

Answer 57

( you can use the easy geometry way to check your answer though )

Answer 58

Yup.

Answer 59

im sorry but can you explain why you did what you did
calc3 student

Answer 60

instructor never explained finding surface area

Answer 61

Well, if I could do that, I'd be done. :) I will as soon as I get through this round, though. Our class didn't do very well on it either. If you want a great discussion of it, though, Khan Acadamy goes into pretty good detail: https://www.khanacademy.org/math/multivariable-calculus/surface-integrals

Answer 62

@camper4834, check this out http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

Answer 63

yep, those are my top two choices to explain just about everything calculus

Answer 64

The Paul's link is great, too.

Answer 65

the only unfortunate thing is it doesn't explain the derivation

Answer 66

For how much gets left out of books and not explained in class, it's a wonder anyone learns anything at all.

Answer 67

true. It gets worse the higher you go

Answer 68

That's encouraging. On the bright side (pun intended), the full moon is just coming up here, and it's *beautiful*. And so are my two daughters, who I'm really starting to miss. I'm going to take a break for dinner, and I'll be back to finish this. Thanks, everyone for your help so far.

Answer 69

But, I hope not to be too lacking when I teach

Answer 70

lol, and yea, calc III sucks for memorization. There are some practice tests you can find online and use

Answer 71

http://www.math.utah.edu/online/2210/Exams/Prac_Final_Answers/Prac_Final_Answers.pdf
https://wumath.wustl.edu/exams/old-exams/math233
http://www.math.jhu.edu/~santhana/202F10/Review/Final.pdf

Just google Calculus III final exam

Answer 72

I hear you @FibonacciChick666. I've never wanted to start teaching again so much as since I've gone back to school. It's like I feel compelled to save the world.

And that's funny...the link you gave me is from my school--I'm at the University of Utah.

Answer 73

I spent 72 hours the week of the final teaching myself the information. I had a very bad graduate student for the class.
lol first link that came up, but then if you can do those, you should be fine. :)

Answer 74

and yea, bad teachers make me want to teach, but common core man.... it is a disgrace

Answer 75

I am still not sure how I can teach effectively with those limitations

Answer 76

I'll never understand the bureaucratic hostility towards an iterative approach to understanding. It works *great* on Khan. Yet my physics dept continues to lose 85% of their declared majors by teaching the topics same way Newton would have learned them (had he not invented them).

Answer 77

.... yep. But don't worry hs graduates won't be able to do algebra 1 in a few years.

Answer 78

let alone physics

Answer 79

Perhaps the problem is precisely that people who are great teachers can't stand the thought of doing it wrong. But I digress. Here's what I have so far:

Answer 80

meh maybe, but anywho, shouldn't your phi(since you bisected the chord) be twice what you got? Argue it if you disagree please

Answer 81

also, what program are you using????? I quite like it

Answer 82

The program is Photoshop. I've been using it for ...well since version 2.0. I'm what you'd call a non-traditional student (in lots of ways).

As for the chord bit, I've no idea.

Answer 83

lol I didn't know photoshop could do that. anyways here is what I am seeing I assume this is your picture, if it is correct?

Answer 84

so then your representative looks like this, but what you have is the second