A cylindrical solid has a cylindrical circular hold drilled out of the center. Bascially, it's a circular cylinder with a hollow spot right down the middle. Find the surface area of the resulting solid.

radius of larger circle: 2in
radius of smaller circle (hollow area): 1in
height: 3in

I got this problem wrong on the test, but here's what I did:
I used the formula for the surface area of a cylinder. I found the surface area for the entire cylinder, which was 20pi. Then found the area of the smaller circles, and got 2pi. My resulting answer was 18pi.

Can someone show me where I went wrong? Thanks!

surface area of two end: PI(4-2)* two ends

surface are of outer : PI*2*3
surface area of inner: PI*1*3

Total area: PI (3+6+2)
check my thinking.

Another question, hehe: By "two ends", you mean the radii of the circles?

The way I looked at was this :

Surface area of outside cylinder = 2pi(2) = 4pi .... 2pi(r)(h)
Surface area of inside of cylinder = 2pi(1) = 2pi

Surface area of top ("washer" shape) = pi(2^2) - pi(1^2) = 3pi

Now the question becomes whether to include the base in the surface area calculation.

Students in the last few days have been using the definition lateral surface area to exclude the base.

So if we include the base the total surface area would be 4pi + 2pi + 6pi = 12pi
if we exclude it then 9pi

Reiny is right on the ends, 3PI for each. I included the base.

Find the surface area of the resulting SOLID.

R = 2, r = 1, h = 3
Outer
Outer cylinder = 2R(Pi)h
Outer cylinder = 2(2)Pi(3) = 12Pi
Inner cylinder = 2r(Pi)h
Inner cylinder = 2(1)Pi(3) = 6Pi
Ends = [PiR^2 - Pi(1^2)]2
Ends = [Pi(2^2)-Pi(1^2)]2 = 6Pi
Total surface area = 24Pi

WHERE DID I GO WRONG, IF I DID??

Lateral surface area of cylinder standing on one end = 2(2)Pi(3) = 12Pi

To find the surface area of the resulting solid, you need to calculate the surface area of the outer cylinder and subtract the surface area of the inner cylinder.

First, let's calculate the surface area of the outer cylinder:
The formula for the surface area of a cylinder is A = 2πrh + 2πr^2, where r is the radius and h is the height.

Given the radius of the larger circle (outer cylinder) is 2 inches and the height is 3 inches, we can plug in these values into the formula:
A_outer = 2π(2)(3) + 2π(2)^2
= 12π + 8π
= 20π

So the surface area of the outer cylinder is 20π square inches.

Next, let's calculate the surface area of the inner cylinder (the hollow area):
The radius of the smaller circle (inner cylinder) is 1 inch and the height is the same as the outer cylinder, 3 inches.

A_inner = 2π(1)(3) + 2π(1)^2
= 6π + 2π
= 8π

Therefore, the surface area of the inner cylinder is 8π square inches.

To find the surface area of the resulting solid, subtract the surface area of the inner cylinder from the surface area of the outer cylinder:
A_result = A_outer - A_inner
= 20π - 8π
= 12π

So, the surface area of the resulting solid is 12π square inches.

In your calculation, you correctly found the surface area of the outer cylinder as 20π, but you made a mistake in finding the surface area of the inner cylinder. You calculated it as 2π, which should have been 8π. Therefore, subtracting the correct value of 8π would result in the correct answer of 12π.