Hello, pakman!

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

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˛.