zakownikbj

Answered

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 $\pi {\displaystyle {\int}_{0}^{1}(\sqrt{x}{)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx.}$.

When I do this, I get $\frac{\pi}{5}$. The answer is $\frac{\pi}{8}$. What am I doing wrong?

Answer & Explanation

Klecanlh

Expert

2022-09-24Added 11 answers

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 $\pi (\sqrt{x}/2{)}^{2}$

- $V=\int \text{"Cross-sectional Area"}\phantom{\rule{thinmathspace}{0ex}}\text{}d\text{"axis perp to cross section"}$

$V=\pi {\int}_{0}^{1}{\left({\displaystyle \frac{\sqrt{x}}{2}}\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=\pi {\int}_{0}^{1}{\displaystyle \frac{x}{4}}\phantom{\rule{thinmathspace}{0ex}}dx={\pi {\displaystyle \frac{{x}^{2}}{8}}|}_{0}^{1}={\displaystyle \frac{\pi}{8}}$

Dymnembalmese2n

Expert

2022-09-25Added 2 answers

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 ($\pi \underset{0}{\overset{1}{\int}}(\sqrt{x}{)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$) should give $\pi /2$ which, multiplied by the new $\frac{1}{4}$ from above, gives the correct answer.

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