A glass vase has the shape of the solid obtained by rotating about the y-axis the area in the first quadrant lying over the x-interval [0,a] and under the graph of \(\displaystyle y=\sqrt{x}\)

Determine how much glass is contained in the vase.

Hi Exotique,

\(\displaystyle y=\sqrt{x}\ \Rightarrow\ x=y^2\)

As the curve is being rotated about the y-axis, we need the limits on the y-axis.

Also, we integrate the discs with radius=x.

\(\displaystyle x=0,\ y=0\)

\(\displaystyle x=a,\ y=\sqrt{a}\)

The volume of the vase is \(\displaystyle \int_{y=0}^{\sqrt{a}}{\pi}x^2dy\)

\(\displaystyle ={\pi}\int_{0}^{\sqrt{a}}y^4dy\)

If the glass is the solid part under the graph above the x-axis, with a hollow interior above the graph,

subtract this integral from \(\displaystyle {\pi}a^2(\sqrt{a})\)