W9.3.8 volume by rotation

[karush]

http://mathhelpboards.com/attachment.php?attachmentid=5579&stc=1
W9.3.8
Let $S$ be the region of the xy-plane bounded above by the curve $x^{3} y = 64$
below by the line $y = 1$, on the left by the line $x = 2$,
and on the right by the line $x = 4$. Find the volume of the solid obtained by rotating S around
(a) the -axis,
(b) the line $y = 1$,
(c) the y-axis,
(d) the line $x = 2$ . Not sure how to set this up

Was going to use shell method which basically is

$$2\pi\int_{a}^{b}y f\left(y\right) \,dy $$

 

[Prove It]

W9.3.8
Let $S$ be the region of the xy-plane bounded above by the curve $x^{3} y = 64$
below by the line $y = 1$, on the left by the line $x = 2$,
and on the right by the line $x = 4$. Find the volume of the solid obtained by rotating S around
(a) the -axis,
(b) the line $y = 1$,
(c) the y-axis,
(d) the line $x = 2$ . Not sure how to set this up

Was going to use shell method which basically is

$$2\pi\int_{a}^{b}y f\left(y\right) \,dy $$

Since you are rotating around the x axis, approximate the area under the curve with rectangles of width $\displaystyle \begin{align*} y = \frac{64}{x^3} \end{align*}$ and very small length $\displaystyle \begin{align*} \Delta x \end{align*}$. Then the volume can be approximated by rotating each of these rectangles around the x axis, giving small cylinders of radius $\displaystyle \begin{align*} y = \frac{64}{x^3} \end{align*}$ and height $\displaystyle \begin{align*} \Delta x \end{align*}$. Then the volume of each small cylinder is $\displaystyle \begin{align*} \pi \, y^2 \, \Delta x \end{align*}$ and the total volume is then approximately $\displaystyle \begin{align*} \sum{ \pi \, y^2 \, \Delta x } \end{align*}$.

As we increase the number of rectangles (to infinity) by making $\displaystyle \begin{align*} \Delta x \end{align*}$ smaller (to 0), the approximation becomes exact and the sum converges on an integral. Thus the volume generated by rotating $\displaystyle \begin{align*} y = \frac{64}{x^3} \end{align*}$ between x = 2 and x = 4 is

$\displaystyle \begin{align*} V &= \int_2^4{ \pi \, \left( \frac{64}{x^3} \right) ^2 \, \mathrm{d}x } \\ &= 4096\,\pi \int_2^4{ x^{-6}\,\mathrm{d}x } \\ &= 4096\,\pi \, \left[ \frac{x^{-5}}{-5} \right]_2^4 \\ &= -\frac{4096}{5}\,\pi\,\left[ \frac{1}{x^5} \right]_2^4 \\ &= -\frac{4096}{5} \, \pi \, \left( \frac{1}{4^5} - \frac{1}{2^5} \right) \\ &= -\frac{4096}{5}\,\pi \, \left( -\frac{31}{1024} \right) \\ &= \frac{124}{5}\,\pi \end{align*}$

Now we have to think about what would happen if we remove the volume of the cylinder that is formed from there being a boundary at y = 1 as well...

 

[karush]

Thanks for all the steps
I'm going to do more volume problems today till I get a good handle on it

 

[karush]

b. About $y=1$

$$y_1=\frac{64}{{x}^{3}}$$

$$\pi\int_{2}^{4}\left(y_1-1\right)^2 \,dy = \frac{74 \pi}{5 }$$

 

[karush]

About y-axis

$$\pi\int_{1}^{8}\left(x_1(y)^2 -2^2 \right) \,dy = 20 \pi$$

About x=2 ?

 

[MarkFL]

b. About $y=1$

$$y_1=\frac{64}{{x}^{3}}$$

$$\pi\int_{2}^{4}\left(y_1-1\right)^2 \,dy = \frac{74 \pi}{5 }$$

Let's check that using the shell method:

\(\displaystyle dV=2\pi rh\,dy\)

\(\displaystyle r=y-1\)

\(\displaystyle h=x-2=2\left(2y^{-\frac{1}{3}}-1\right)\)

And so we have:

\(\displaystyle dV=4\pi (y-1)\left(2y^{-\frac{1}{3}}-1\right)\,dy\)

\(\displaystyle V=4\pi\int_1^8 2y^{\frac{2}{3}}-y-2y^{-\frac{1}{3}}+1\,dy=\frac{74\pi}{5}\checkmark\)

 

[karush]

Well that helped a lot to understand the shell mdthod

 

[MarkFL]

About y-axis

$$\pi\int_{1}^{8}\left(x_1(y)^2 -2^2 \right) \,dy = 20 \pi$$

About x=2 ?

About the $y$-axis:

Washer method:

\(\displaystyle dV=\pi\left(R^2-r^2\right)\,dy\)

\(\displaystyle R=x\)

\(\displaystyle r=2\)

Hence:

\(\displaystyle dV=\pi\left(x^2-2^2\right)\,dy=4\pi\left(4y^{-\frac{2}{3}}-1\right)\,dy\)

And so we have:

\(\displaystyle V=4\pi\int_1^8 4y^{-\frac{2}{3}}-1\,dy=20\pi\)

Shell method:

\(\displaystyle dV=2\pi rh\,dx\)

\(\displaystyle r=x\)

\(\displaystyle h=y-1=64x^{-3}-1\)

Hence:

\(\displaystyle V=2\pi\int_2^4 64x^{-2}-x\,dx=20\pi\)

Can you give $x=2$ as the axis of revolution a try now? Whichever way you do it, I will do it the other way as a check. :)

 

[karush]

$y_1=\frac{64}{{x}^{3}}$

$x-2=r$

$$2 \pi \int_{2 }^{4 }r\left(y_1-1\right) \,dx=4\pi$$

 

[MarkFL]

$y_1=\frac{64}{{x}^{3}}$

$x-2=r$

$$2 \pi \int_{2 }^{4 }r\left(y_1-1\right) \,dx=4\pi$$

Using the disk method:

\(\displaystyle dV=\pi r^2\,dy\)

\(\displaystyle r=x-2=2\left(2y^{-\frac{1}{3}}-1\right)\)

Hence:

\(\displaystyle V=4\pi\int_1^8 4y^{-\frac{2}{3}}-4y^{-\frac{1}{3}}+1\,dy=4\pi\)