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{\bf Question}

Sketch the region enclosed by the given curves and find the volume
of the solid generated when it is revolved about the $x$-axis:

$y^2=x$, $y=1$, $x=0$.

{\bf Answer}


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Either use washer method:

$$\int_{x=0}^{x=1} \left \{\pi(1)^2-\pi(\sqrt x)^2 \right\} \,dx =
\pi \int_{x=0}^{x=1}(1-x) \,dx = \pi \left[x-\frac{x^2}{2}
\right]_0^1 = \frac{\pi}{2} $$

or use shell method:

$$ \int_{y=0}^{y=1} (2\pi y)(y^2) \,dy = 2\pi \int_{y=0}^{y=1}y^3
\,dy = 2\pi \left[\frac{y^4}{4} \right]_0^1 = \frac{\pi}{2} $$


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