Rotating an ellipse

Calculus Level 3

( x a ) 2 + ( y b ) 2 = 1 \left(\dfrac{x}{a}\right)^2 + \left(\dfrac{y}{b}\right)^2=1

An ellipse is described by the equation above. What is ithe volume of the solid that is obtained by rotating the ellipse around the x x -axis?

V = 3 2 π ( 2 a + b 3 ) 3 V=\frac{3}{2}\pi \left(\frac{2a+b}{3}\right)^3 V = 3 2 π π a b 2 V=\frac{3}{2}\pi\pi ab^2 V = 4 3 π a b 2 V=\frac{4}{3}\pi ab^2 V = 3 2 π π a 2 b V=\frac{3}{2}\pi\pi a^2b V = 4 3 π a 2 b V=\frac{4}{3}\pi a^2b

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1 solution

Poca Poca
May 2, 2018

The ellipse intersects with the x-axis in P ( a ; 0 ) P(a;0) and Q ( a ; 0 ) Q(-a;0) , which can be obtained from the above equation for y = 0 y=0 . We have ( x a ) 2 + ( y b ) 2 = 1 \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2=1 , which is equivalent to y = ± b 1 x 2 a 2 y=\pm b\sqrt{1-\frac{x^2}{a^2}} . For the part of the ellipse that lies above the x-axis the sign will be + + . Now let f ( x ) = b 1 x 2 a 2 f(x)= b\sqrt{1-\frac{x^2}{a^2}} , which implies [ f ( x ) ] 2 = b 2 ( 1 x 2 a 2 ) [f(x)]^2= b^2(1-\frac{x^2}{a^2}) . Hence by the application of the disk method , the volume is:

V = π a a ( b 2 x 2 a 2 ) d x V=\pi \int_{-a}^a{\left(b^2-\frac{x^2}{a^2}\right)dx}

V = 2 π 0 a ( b 2 x 2 a 2 ) d x V= 2\pi \int_{0}^a{\left(b^2-\frac{x^2}{a^2}\right)dx}

V = 2 π [ b 2 x + b 2 x 3 3 a 2 ] 0 a V= 2\pi \left[b^2x+\frac{b^2x^3}{3a^2} \right]_0^a

V = 2 π ( b 2 a + b 2 a 3 3 a 2 ) V= 2\pi \left(b^2a+\frac{b^2a^3}{3a^2} \right)

V = 4 3 π a b 2 V= \frac{4}{3}\pi ab^2

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