MHB Volumn of x=2sqrt y. About y axis

[karush]

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 $162\pi$ ?

 

[MarkFL]

The book is correct...you have:

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

 

[karush]

OK, I see it.