Question

find the surface area of the part of the cone z=sqrt(x^2+y^2) that lies between the plane y=x and cylinder y = x^2

Answer 1

Have you already tried parametrizing your surface with polar coordinates (cylindrical specifically)?

Answer 2

I don't really understand this stuff. Teacher makes mistakes in his examples of it which just confuses me.I honestly don't know where to start.

Answer 3

.....I did try one method kept getting the double integral of sqrt(2) and had no idea what the region I was integrating on was.

Answer 4

Sorry, but I don't think that the graph of y=x^2 rotated around the y-axis is a cylinder...

Answer 5

Sounds like you're attached to circular cylinders, Ti. \(y=x^2\) is a perfectly good parabolic cylinder.

Answer 6

its a 3d graph, not a rotation

Answer 7

I'd recommend using the first formula on this page:
http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

and then trying a polar transform.

Answer 8

Slaw I tried parametrizing with x=x y=x and x=f(x)=sart(x^2+y^2) and used the formula
\[\int\limits_{?}^{?}\int\limits_{?}^{?}\sqrt{1+f _{x}^2+f_{y}^2}dA\]

Answer 9

when I got the double integral of sqrt 2

Answer 10

whats the region D then?...are they from x^2 to x and 0 to 1?

Answer 11

I don't trust myself here cause I don't fully understand the concept.....

Answer 12

It's the region in the \(x,y\)-plane bounded by those two curves.

Answer 13

so its
\[\int\limits\limits_{0}^{1}\int\limits\limits_{x^2}^{x}\sqrt{1+f _{x}^2+f_{y}^2}dA\]
in this case?

Answer 14

More formally,
\[D=\{(x,y)|0 \leq x\leq 1, x \leq y \leq x^2\}.\]
Which gives rise to precisely the bounds you have, yes.

Answer 15

thank you.....its hard to tell if I'm doing it right when the professor screws up the boundaries in class

Answer 16

I know the feeling. Try deriving that formula using a parametrization; it really helped me understand it when I first learned this stuff.