Solution to Math 252 Quiz 3: Finding the Volume of a Rotated Washer, Quizzes of Calculus

The solution to quiz 3 question in math 252, which asks to find the volume of a solid obtained by rotating a washer around the y-axis. The washer's cross-section is a circle with inner radius x+1 and outer radius 2x+1. The document calculates the area of the cross-section and integrates it from x=0 to x=1 to find the volume.

Typology: Quizzes

Pre 2010

Uploaded on 07/29/2009

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Math 252 Quiz 3 Solution
1. (10pts) Find the volume of the solid obtained by rotating the shaded region below
about the line y=1. [Be sure to simplify your answer.]
Solution.
A cross section perpendicular to the x-axis at position x(0 x1) is a washer with
inner radius x+ 1 and outer radius 2x+ 1 (see the picture above). Hence the area of
that cross section is
A(x) = π(2x+ 1)2π(x+ 1)2=π(4x2+ 4x+ 1 (x2+ 2x+ 1)) = π(3x2+ 2x)
Therefore the volume of the solid is
Z1
0
A(x)dx =πZ1
0
(3x2+ 2x)dx =π[x3+x2]1
0=π(1 + 1 00) = 2π.

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Math 252 Quiz 3 Solution

  1. (10pts) Find the volume of the solid obtained by rotating the shaded region below about the line y = −1. [Be sure to simplify your answer.] Solution.

A cross section perpendicular to the x-axis at position x (0 ≤ x ≤ 1) is a washer with inner radius x + 1 and outer radius 2x + 1 (see the picture above). Hence the area of that cross section is

A(x) = π(2x + 1)^2 − π(x + 1)^2 = π(4x^2 + 4x + 1 − (x^2 + 2x + 1)) = π(3x^2 + 2x)

Therefore the volume of the solid is ∫ (^1)

0

A(x)dx = π

0

(3x^2 + 2x)dx = π[x^3 + x^2 ]^10 = π(1 + 1 − 0 − 0) = 2π.