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Question 58 of 96

Q.Volume of the solid obtained by revolving the area of the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 about major and minor axes are in the ratio :

(a) b2:a2b^2 : a^2
(b) a2:b2a^2 : b^2
(c) a:ba : b
(d) b:ab : a
Puducherry TnboardTamil Nadu HSC (DGE) Board 2017MCQ· 1mImportance★★★★★
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Revolving the ellipse about its major axis gives volume 43πab2\frac43\pi ab^2, and about its minor axis gives 43πa2b\frac43\pi a^2b; their ratio is b:ab:a.

  1. Ellipse: x2a2+y2b2=1⇒y2=b2(1−x2a2)\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \Rightarrow y^2=b^2\left(1-\dfrac{x^2}{a^2}\right).
  2. Volume when revolved about the major axis (xx-axis), by the disk method: V1=π∫−aay2 dx=πb2[x−x33a2]−aa=πb2(2a−2a3)=43πab2V_1=\pi\displaystyle\int_{-a}^{a}y^2\,dx=\pi b^2\left[x-\dfrac{x^3}{3a^2}\right]_{-a}^{a}=\pi b^2\left(2a-\dfrac{2a}{3}\right)=\dfrac{4}{3}\pi ab^2. …

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