zakownikbj

2022-09-23

Finding volume of solid
Suppose that a solid is formed in such a way that each cross section perpendicular to the x-axis, for $0\le x\le 1$, is a disk, a diameter of which goes from the x-axis out to the curve $y=\sqrt{x}$.
Find the volume of the solid.
For this I use the disk formula. So $\pi {\int }_{0}^{1}\left(\sqrt{x}{\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx.$.
When I do this, I get $\frac{\pi }{5}$. The answer is $\frac{\pi }{8}$. What am I doing wrong?

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Klecanlh

Expert

Explanation:
You are told that the cross-sections perpendicular to the x-axis are disks where the diameter is from $y=\sqrt{x}$ to $y=0$.
- The diameter of this disk is $\sqrt{x}$
- The radius of this disk is $\sqrt{x}/2$
- The cross-sectional area is $\pi \left(\sqrt{x}/2{\right)}^{2}$
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$V=\pi {\int }_{0}^{1}{\left(\frac{\sqrt{x}}{2}\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=\pi {\int }_{0}^{1}\frac{x}{4}\phantom{\rule{thinmathspace}{0ex}}dx={\pi \frac{{x}^{2}}{8}|}_{0}^{1}=\frac{\pi }{8}$

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Dymnembalmese2n

Expert

Step 1
You're doing two things wrong.
One, it is explicitly saying that the diameter of each shell is $\sqrt{x}$, so the radius would be $\sqrt{x}/2$
Step 2
Next, you're integrating wrong. The integral you have ($\pi \underset{0}{\overset{1}{\int }}\left(\sqrt{x}{\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$) should give $\pi /2$ which, multiplied by the new $\frac{1}{4}$ from above, gives the correct answer.

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