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.
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 $\displaystyle {\pi}a^2(\sqrt{a})$
Yes,
if the vase really is that shape.
We get the same answer by integrating the surface areas of the washers
(denoted by the 2 red discs).
This is equivalent to subtracting the interior volume (of air)
from the volume of a cylinder.
$\displaystyle \int_{0}^{\sqrt{a}}\left({\pi}a^2-{\pi}x^2\right)dy={\pi}\int_{0}^{\sqrt{a}}\left(a^ 2-x^2\right)dy$
$\displaystyle ={\pi}\int_{0}^{\sqrt{a}}\left(a^2-y^4\right)dy={\pi}(\sqrt{a})^5-{\pi}\frac{(\sqrt{a})^5}{5}$