• Use cylindrical shells to find the volume of the solid obtained by rotating about the $x$-axis the region under curve $y=\sqrt {x}$ from $0$ to $1$.

Shell radius= $y$

Shell height= $1-y^2$

$V=2\pi \int _{0}^{1}y\left( 1-y^{2}\right) dy$.

I didn't understand the shell height, why it is $1-y^2$? Why not $y^2$? I think, shell height should be $y^2$. Can you explain?


The region is bounded on the left by $y=\sqrt{x}$, which can be rewritten as $x=y^2$.

The region is also bounded on the right by $x=1$

Hence $$1-y^2$$


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