# Find the area of a surface (double integral)

• Feb 13th 2007, 06:38 AM
pakman
Find the area of a surface (double integral)
I am having some trouble with this problem. It states that I must find the area of the indicated surface of:

The part of the cylinder x^2 + x^2 = 9 that is directly over the rectangle in the xy-plane with vertices (0,0), (2,0), (2,3), and (0,3).

I'm still kind of sketchy on how to draw 3-dimensional objects on an xyz-plane, but it's obviously fairly obvious how to draw the rectangle given the points. However, how exactly do I know what part of the cylinder is within that rectangle? Plus, the formula to solve this kind of problem, I think is...

S = integral

A(G) = SS sqrt(fx^2 + fy^2 + 1) dxdy

There is no y in this function, what do I do? Thanks.
• Feb 13th 2007, 10:40 AM
ThePerfectHacker
Quote:

Originally Posted by pakman
The part of the cylinder x^2 + x^2 = 9 that is directly over the rectangle in the xy-plane with vertices (0,0), (2,0), (2,3), and (0,3).

What type of cyclinder is this :confused: ?

Do you mean,
x^2+z^2=9

If that is what you mean then,
z^2=9-x^2
Note it says "over" thus you are looking at the upper part.
Thus,
z=\sqrt(9-x^2) is positive.

Then,
z_x is found the regular way.
And relative to the y variable this is constant, thus,
z_y is zero.
• Feb 13th 2007, 10:41 AM
ticbol
Quote:

Originally Posted by pakman
I am having some trouble with this problem. It states that I must find the area of the indicated surface of:

The part of the cylinder x^2 + x^2 = 9 that is directly over the rectangle in the xy-plane with vertices (0,0), (2,0), (2,3), and (0,3).

I'm still kind of sketchy on how to draw 3-dimensional objects on an xyz-plane, but it's obviously fairly obvious how to draw the rectangle given the points. However, how exactly do I know what part of the cylinder is within that rectangle? Plus, the formula to solve this kind of problem, I think is...

S = integral

A(G) = SS sqrt(fx^2 + fy^2 + 1) dxdy

There is no y in this function, what do I do? Thanks.

I assume the x^2 +x^2 = 9 is a typo. Following your question here, I see that that should be x^2 +z^2 = 9, a cylinder in x and z.

The figure is like a loaf of bread.

Since the base is in the xy plane, then the cylinder x^2 +z^2 = 9 must be converted to a function of x and y---which is z here.
x^2 +z^2 = 9
z^2 = 9 -x^2
z = f(x,y) = sqrt(9 -x^2) -----------(i)

The surface area I assume you want to find is the curved surface of the loaf of bread, the portion of the cylinder over the rectangular base.

Yes, the double integral is in the form
A = SSsqrt[(partial derivative of z with respect to x)^2 +(partial derivative of z with respect to y)^2 +1](dx dy)
So,
dA = sqrt[(-x /sqrt(9 -x^2))^2 +0 +1](dx dy) ------y has no part in the z = sq(9 -x^2)
dA = sqrt[(x^2)/(9 -x^2) +1](dx dy) -------------**

A = SS{sqrt[((x^2)/(9-x^2)) +1]}(dx dy) -------------(ii)

The boundaries of dx are from x=0 to x=2.
The boundaries of dy are from y=0 to y=3

You can do it now.
(Sorry, I don't have enough time to finish it now. Job time. If this were tonight, I'd finish it.)
• Feb 13th 2007, 01:44 PM
Soroban
Hello, pakman!

I assume there's a typo: you didn't mean to type x˛ twice.

Quote:

Find the area of the surface of the part of the cylinder x˛ + z˛ = 9 that is
directly over the rectangle in the xy-plane with vertices (0,0), (2,0), (2,3), and (0,3).

As far as I know, no one can "draw 3-dimensional objects on an xyz-plane".
The best we can do is try to visualize the solid.

The cylinder x˛ + z˛ .= .9 has a radius of 3 and is centered on the y-axis.
. . On a standard xyz-graph, it runs left-and-right.

On the xy-plane (the "floor"), we have that 2-by-3 rectangle.
And we want the portion of the cylinder directly above the rectangle.

The solid looks like a slice of cake . . . something like this:
Code:

            *----------------*         *::::::::::::::::*  |       *::::::::::::::::*    |       *----------------*      |       \                \    |         \                \    |3         \                \  |           \                \  |           \                \θ|             \                \|             *----------------*                     3
Ha! . . . We can find that area without Calculus.

The lateral area of the entire cylinder (about that rectangle)
. . is: .A .= .2πrh .= .2π(3)(3) .= .18π units˛.

We want a fraction of that area . . . what fraction?
The ratio is the central angle θ to 360°.

It can be shown that: .tan θ = 2/3 . . θ ≈ 33.69°

Therefore: .S .= .(33.69/360) x 18π . .5.29 units˛.