Question

what is the value of the area of part of the surface z=y^2-x^2 bounded by the cylinders x^2+y^2=1 and x^2+y^2=4, help

Answer 1

I think it should be
\[\int\limits_{1}^{2} \int\limits_{y=\sqrt{1-x ^{2}}}^{y=\sqrt{4-x ^{2}}} y ^{2} -x ^{2} dydx\]

Answer 2

@mukushla Does that look right?

Answer 3

i need to understand how we are getting the limits

Answer 4

surface of the inner cylinder to the surface of the outer cylinder

Answer 5

where did the x limits 1 & 2 come from ?

Answer 6

when y=0 x=1 on the inner cylinder and when y=0 x=2 on the outer cylinder

Answer 7

thank you Algebraic i was stuck!

Answer 8

No problem... did you calculate it and get an answer?

Answer 9

that's what am working on now

Answer 10

I'm not so sure that's right algebraic

Answer 11

surface area over a region D in the xy-plane for a function \(z=f(x,y)\) is given by\[S=\iint\limits_D\sqrt{[f_x]^2+[f_y]^2+1}dA\]

Answer 12

gotta take some partial derivatives

Answer 13

I would do this sucker in cylindrical coordinates too, the integral will be easier

Answer 14

but the limits @ Algebraic are correct right ?

Answer 15

yes, but the integrand is wrong, and integrating that is gonna suck; you should put it in cylindrical coordinates

Answer 16

actually the outer bounds are right, not sure about the inner

Answer 17

can you explain a little further on how we are getting the limits

Answer 18

It makes much more sense in cylindrical coordinates, first sketch the region D

Answer 19

D is the region on the xy-plane between two cylinders r=1 and r=2, so those are our limits for the radius
since it is a circular region the limits on theta will be from 0 to 2pi
make sense now?

Answer 20

whoops, I did volume. Didn't read closely...

Answer 21

@KANNYTE hello?

Answer 22

its so beautiful to be around geneouses, am begining to understand now

Answer 23

so what is the integrand ?(leave it cartesian for now, you will see that the problem will simplify when we switch to cylindrical)

Answer 24

Turning Test you are making my life so much easier right now

Answer 25

happy to help, multivariable calc is kind of a strong point of mine :)\[\sqrt{[f_x]^2+[f_y]^2+1}=?\]

Answer 26

every time I said "cylindrical coordinates" I meant "polar" btw :P
cylindrical is only for integrating volumes and such...

Answer 27

am actually writting notes from you right now

Answer 28

here's a good site for this kind of stuff:
http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

Answer 29

well, when you're done taking notes please tell me what you get for the integrand\[\sqrt{[f_x]^2+[f_y]^2+1}\]

Answer 30

okay am working on it

Answer 31

can the y limits be 1 & 2 instead of x since the radius of the cylinders is also extending along the y axis ?

Answer 32

1 and 2 will not be limits of y; they will be the limits of the radius r when we switch to polar coordinates, which we don't need to do just yet
for now just do what I said above and take the partial derivatives in cartesian coordinates to find the integrand

Answer 33

what is\[f_x=\frac{\partial z}{\partial x}\]?

Answer 34

-2x

Answer 35

and what is \(f_y\) ?

Answer 36

2y

Answer 37

yes
(btw the cylinders do not extend along y, they extend along z)

Answer 38

this graph is in the xy-plane, the cylinders extend out of the page