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Math - Geometry

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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!

  • Math - Geometry -

    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.

  • Math - Geometry -

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

  • Math - Geometry -

    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

  • Math - Geometry -

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

  • Math - Geometry -

    Find the surface area of the resulting SOLID.

    R = 2, r = 1, h = 3
    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


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

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