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**dgiroux48****Member**- Registered: 2013-04-18
- Posts: 4

Having trouble even figuring out where to start on this problem. Our professor uploaded it on a kind of "cumulative final study guide", but I think we either use change of variables or surface integrals/parameterized surfaces (two topics we have covered). Let me know if you can help me tackle this one! Thanks!

2. We wish to build a fence on the landscape z = −1 − 8x^2 − 16y^2 from ground level up to height z = 0. The path of the fence is y=x^2 as x varies from x=−2 to x=+2. What will be the area of (the face of) the fence?

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

there is a formula for the length of a curve that might help. jane fairfax gave it to me on this forum a few years ago.

search for length of curve or something... then put that together with the height witha product i guess and maybe it'll work.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

I searched for Jane's formula but couldn't find it.

but i did find this weird discussion of mine and carlos.

http://www.mathisfunforum.com/viewtopic.php?id=2168

**igloo** **myrtilles** **fourmis**

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**dumbcow****Guest**

The surface integral would give the area of a specified surfaced of the paraboloid (z) but we want area of the fence which comes up from surface to z=0 along the points y=x^2.

The area of fence can be found by multiplying height by length of fence dictated by path (y=x^2) along surface

Area = integral [-2,2] |z| dL

dL^2 = dx^2 + dy^2 +dz^2

hence

dL = sqrt(1 + (dy/dx)^2 + (dz/dx)^2 ) dx

since path of fence is y=x^2, we can put z in terms of x

z = -1 -8x^2 -16x^4

dy/dx = 2x

dz/dx = -16x -64x^3

substituting into Area integral:

Area = integral[-2,2] (1+8x^2+16x^4) sqrt(1+260x^2+2048x^4 +4096x^6) dx

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