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Find the volume when a curve $16x^2-(y-8)^2=32$ is rotated completely about $y$ axis between $y=0$ and $y=16$ Can anyone help me with this question and a little sketching of the graph might helped.

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    $\begingroup$ If you are having a hard time sketching the graph, can you at least identify the object being graphed (i.e., conic section)? $\endgroup$
    – Ron Gordon
    Jan 10 '16 at 14:05
  • $\begingroup$ I got no idea about that $\endgroup$
    – Ameet
    Jan 10 '16 at 14:10
  • $\begingroup$ Well, no offense, but maybe this problem is beyond your capability. $\endgroup$
    – Ron Gordon
    Jan 10 '16 at 14:11
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Your curve looks like this.

The volume is $V=\int_0^{16}{\pi (x(y))²dy}$.

$$\begin{align} 16x²&=32+(y-8)²\\ x²&=2+\frac{1}{16}(y-8)²\\ V&=\int_0^{16}{\pi (2+\frac{1}{16}(y-8)²)dy}\\ &=\int_0^{16}{\pi(2+\frac{1}{16}(y-8)²)dy}\\ \text{Let } u&=y-8\\ V&=\int_{-8}^{8}{\pi(2+\frac{1}{16}u²)du} \end{align}$$ I think you can take it from here.

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