odcinaknr

2022-09-07

Volume of revolved solid using shell method: finding height

The problem that I am working with is:

Find the volume of the solid of revolution formed by rotating the region R bounded by $y=4+{x}^{2},\phantom{\rule{thickmathspace}{0ex}}x=0,\phantom{\rule{thickmathspace}{0ex}}y=0,\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}x=1$ about the line $y=10$.

I have the following so far (using the shell method):

$V={\int}_{a}^{b}2\pi \phantom{\rule{thinmathspace}{0ex}}r\phantom{\rule{thinmathspace}{0ex}}h\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{0ex}{0ex}}r=10-y\phantom{\rule{0ex}{0ex}}c=2\pi (10-y)\phantom{\rule{0ex}{0ex}}h=?$

$V=2\pi {\int}_{0}^{1}(10-y)hdy$

I've been searching for an example in my book where there is a horizontal rotation that is not over the x-axis, but I have had no luck.

Would the height just be $\sqrt{y-4}$ or would it be $10-\sqrt{y-4}$ and why?

The problem that I am working with is:

Find the volume of the solid of revolution formed by rotating the region R bounded by $y=4+{x}^{2},\phantom{\rule{thickmathspace}{0ex}}x=0,\phantom{\rule{thickmathspace}{0ex}}y=0,\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}x=1$ about the line $y=10$.

I have the following so far (using the shell method):

$V={\int}_{a}^{b}2\pi \phantom{\rule{thinmathspace}{0ex}}r\phantom{\rule{thinmathspace}{0ex}}h\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{0ex}{0ex}}r=10-y\phantom{\rule{0ex}{0ex}}c=2\pi (10-y)\phantom{\rule{0ex}{0ex}}h=?$

$V=2\pi {\int}_{0}^{1}(10-y)hdy$

I've been searching for an example in my book where there is a horizontal rotation that is not over the x-axis, but I have had no luck.

Would the height just be $\sqrt{y-4}$ or would it be $10-\sqrt{y-4}$ and why?

pereishen9g

Beginner2022-09-08Added 7 answers

Step 1

We will assume that you really want to use the Shell Method. In that case, we will look at a thin strip of width "dy" at height y, and rotate it about $y=10$.

The complication is that the thin strip has a different shape from $y=0$ to $y=4$ than it does from $y=4$ to $y=5$.

From $y=0$ to $y=4$, the radius is $10-y$, and the "height" of the cylindrical shell is 1. Thus the volume obtained by rotating that part is ${\int}_{y=0}^{4}2\pi (10-y)(1)\phantom{\rule{thinmathspace}{0ex}}dy.$

Step 2

From $y=4$ to $y=5$, the radius is still $10-y$, but the "height" of the cylindrical shell is $1-x$, that is, $1-\sqrt{y-4}$. The required integral is

${\int}_{y=4}^{5}2\pi (10-y)(1-\sqrt{y-4})\phantom{\rule{thinmathspace}{0ex}}dy.$

Calculate the two integrals, and add.

Remark: For this problem, the Method of Washers is less work. Take a cross-section "at" x perpendicular to the x-axis. The outer radius is 10, and the inner radius is ${\int}_{y=4}^{5}2\pi (10-y)(1-\sqrt{y-4})\phantom{\rule{thinmathspace}{0ex}}dy.$

We will assume that you really want to use the Shell Method. In that case, we will look at a thin strip of width "dy" at height y, and rotate it about $y=10$.

The complication is that the thin strip has a different shape from $y=0$ to $y=4$ than it does from $y=4$ to $y=5$.

From $y=0$ to $y=4$, the radius is $10-y$, and the "height" of the cylindrical shell is 1. Thus the volume obtained by rotating that part is ${\int}_{y=0}^{4}2\pi (10-y)(1)\phantom{\rule{thinmathspace}{0ex}}dy.$

Step 2

From $y=4$ to $y=5$, the radius is still $10-y$, but the "height" of the cylindrical shell is $1-x$, that is, $1-\sqrt{y-4}$. The required integral is

${\int}_{y=4}^{5}2\pi (10-y)(1-\sqrt{y-4})\phantom{\rule{thinmathspace}{0ex}}dy.$

Calculate the two integrals, and add.

Remark: For this problem, the Method of Washers is less work. Take a cross-section "at" x perpendicular to the x-axis. The outer radius is 10, and the inner radius is ${\int}_{y=4}^{5}2\pi (10-y)(1-\sqrt{y-4})\phantom{\rule{thinmathspace}{0ex}}dy.$

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