Volumn of x=2sqrt y. About y axis

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SUMMARY

The volume bounded by the curves defined by \(x=2\sqrt{y}\), \(x=0\), and \(y=9\) about the y-axis is calculated using the formula \(V=4\pi\int_0^9 y\,dy\). The correct evaluation of this integral yields a volume of \(162\pi\), confirming the book's answer. The initial incorrect calculation of \(\pi\int_{0}^{9}(2\sqrt{y})^2 \,dy\) resulting in \(160\pi\) was due to a misunderstanding of the volume formula for revolution about the y-axis.

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karush
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Find volume bounded by
$$x=2\sqrt{y},\ \ x=0, \ \ y=9 $$
About the y axis

So..

$$\pi\int_{0}^{9}(2\sqrt{y})^2 \,dy = 160\pi$$

But the book answer is $162\pi$ ?
 
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The book is correct...you have:

$$V=4\pi\int_0^9 y\,dy=2\pi\left(9^2-0^2\right)=162\pi$$
 
OK, I see it.
 

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