Consider the region bounded by the ellipse and the -axis for . Given that , if is the length of the semi-minor axis that minimizes the volume of the solid obtained by revolving about the line , find
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The axis of revolution isn't perpendicular to the region in question, so using the method of shells/disks can be really hard. Using the first Pappus-Guldinus theorem can allow us to compute the volume in terms of a and b easily.
The centroid of R lies on the y -axis (by symmetry), while the y − coordinate is
y ˉ = A ∫ 0 b y d A = A ∫ 0 b y L ( y ) d y
using the formula for the centroid of thin a plate ( A is the area of the region). Since x = ± b a b 2 − y 2 , we have L ( y ) = b 2 a b 2 − y 2 . Therefore,
y ˉ = 2 1 π a b ∫ 0 b b 2 a y b 2 − y 2 = 2 1 π a b 3 b − 2 a ( b 2 − y 2 ) 3 / 2 ∣ ∣ ∣ ∣ 0 b = 3 2 a b 2 ⋅ π a b 2 = 3 π 4 b
Therefore, the centroid lies at ( 0 , 3 π 4 b ) . From Pappus' theorem we know that
V = 2 π A ρ = π 2 a b ρ = 3 2 π 2 ρ
where ρ is the distance of the centroid from the axis of revolution. Hence, we need to find the distance between ( 0 , 3 π 4 b ) and the line y = x − a . The equation of the line perpendicular to y = x − a and passing through ( 0 , 3 π 4 b ) is y = − x + 3 π 4 b . These lines intersect at
x − a = − x + 3 π 4 b ⇒ x = 2 a + 3 π 2 b ⇒ ( 2 a + 3 π 2 b , 2 − a + 3 π 2 b )
∴ ρ = ( 2 a + 3 π 2 b ) 2 + ( 2 a + 3 π 2 b ) 2 = 2 a + 3 π 2 2 b = b 3 + 3 π 2 2 b
Therefore V ( b ) = 3 2 π 2 ( b 3 + 3 π 2 2 b ) . Minimizing by taking the derivative yields
b 2 − 3 + 3 π 2 2 = 0 ⇒ b min = 2 3 π 2 ⇒ ⌈ 1 0 0 0 b min ⌉ = 3 1 6 2