Question

The volume of the solid obtained by revolving the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ about the minor axis is

• A. $48\pi$
• B. $64\pi$
• C. $32\pi$
• D. $128\pi$

Answer 1

Answer: (B)

$64\pi$

The solid will rotate about $x$ axis , the length of solid on $x$ axis is $6$ and the length of solid on $y$ axis is $8$

Volume of solid obtained is ${\int }_{-3}^3\pi y^{2}dx$

From the given equation , we get $y^2=16(1-\frac{x^2}{9})$

Therefore the volume of solid is $\frac{16\pi }{9}{\int }_{-3}^3(9-x^2)dx=\frac{16\pi }{9}(54-18)=64\pi$