The lemniscates r^2=a^2 cos 2 theta revolves about a tangent at the pole. What is the volume generated by it?

temeljil4l

temeljil4l

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2022-08-22

Volume generated by lemniscate revolving about a tangent at the pole.
The lemniscates r 2 = a 2 cos 2 θ revolves about a tangent at the pole. What is the volume generated by it ?
Please explain in detail. I found a couple of answers on finding surface areas, volumes but didn't understood it.
Can someone solve this. I'm stuck in the middle with integrations.

Answer & Explanation

penzionekta

penzionekta

Beginner2022-08-23Added 8 answers

Step 1
The lemniscate with a = 1 is the locus of points (x, y) for which:
( x 2 + y 2 ) 2 = x 2 y 2
and the tangents in the origin are the lines y = ± x. Let R be the region of the x 0 halfplane bounded by the lemniscate. We just need to compute the area A of R, the centroid G of R, then apply the second Pappus' centroid theorem. By using polar coordinates we have:
(1) A = 1 2 π / 4 π / 4 r 2 d θ = 1 2 .
Step 2
By symmetry, the centroid of R lies on the y = 0 line. Its abscissa is given by:
G x = 0 1 x f ( x ) d x 0 1 f ( x ) d x = 0 1 x f ( x ) d x A / 2 = 4 0 1 x f ( x ) d x = 2 2 0 1 x 1 2 x 2 + 1 + 8 x 2 d x (2) = π 4 2
hence the distance of the centroid of R from a tangent line in the origin is just π / 8, and:
(3) V = 2 2 π π 8 A = π 2 4 ,
so, in the general case:
V = 1 4 π 2 | a | 3 .

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