Voulme by integration

Apr 2010
18
3
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.
 
Dec 2009
3,120
1,342
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 \(\displaystyle {\pi}a^2(\sqrt{a})\)
 
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Apr 2010
18
3
The value I obtained was
\(\displaystyle \pi\sqrt{a}^5-\frac{1}{5}\pi\sqrt{a}^5\)
\(\displaystyle =\frac{4\pi}{5}\cdot\sqrt{a}^5\)

Is this correct?
 
Dec 2009
3,120
1,342
The value I obtained was
\(\displaystyle \pi\sqrt{a}^5-\frac{1}{5}\pi\sqrt{a}^5\)
\(\displaystyle =\frac{4\pi}{5}\cdot\sqrt{a}^5\)

Is this correct?
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.

\(\displaystyle \int_{0}^{\sqrt{a}}\left({\pi}a^2-{\pi}x^2\right)dy={\pi}\int_{0}^{\sqrt{a}}\left(a^2-x^2\right)dy\)

\(\displaystyle ={\pi}\int_{0}^{\sqrt{a}}\left(a^2-y^4\right)dy={\pi}(\sqrt{a})^5-{\pi}\frac{(\sqrt{a})^5}{5}\)
 

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