MHB Solids of Rotation bounded by y=bx^(1/2)

[Dethrone]

For $a>0$, let $V$ be the volume created by revolving the region bounded by $y=b\sqrt{x}$ and $x=a$ around the axis $x=a$. The units of $x$, $y$, and $a$ are $[m]$. The units of $b$ are $[m^{1/2}]$. The value of $b$ remains fixed.

For fun and practice, I did this question both with cylindrical and disk method. Also, I'm going to ignore the units for now.

Disk method:

$$dV=\pi \left((a-\left(\frac{y}{b}\right)^2\right)^2dy$$$$V=\pi \int_{0}^{b\sqrt{a}} \left((a-\left(\frac{y}{b}\right)^2\right)^2\,dy$$$$V=\pi a^{5/2}b\left(1-\frac{2b}{3}+\frac{b^3}{5}\right)$$

Cylindrical Shell method:

$$dV=2\pi (a-x)(b\sqrt{x})dx$$
$$V=2\pi \int_{0}^{a} (a-x)(b\sqrt{x})\,dx$$
$$V=\frac{6}{5}\pi a^{5/2}b$$

Anyone find my mistake?

 

[MarkFL]

Your integrals are correctly set up in both cases. When I evaluate them, I get the same result, which is different from both results you obtained.

 

[Dethrone]

Disk method:

I am treating $a$ and $b$ as constants.

$$V=\pi \int_{0}^{b\sqrt{a}} a^2y-\frac{2a}{3b}y^3+\frac{y^5}{5b}\,dy$$
$$V=\pi \left(a^2\sqrt{a}b-\frac{2}{3}a\frac{b^3a\sqrt{a}}{b}+\frac{b^5a^2\sqrt{a}}{5b}\right)\,$$
$$V=\pi \left(a^{5/2}b-\frac{2}{3}a^{5/2}b^2+\frac{b^4}{5}a^{5/2}\right)$$

Please correct me :D

 

[MarkFL]

Check your expansion of:

$$\left(a-\frac{y^2}{b^2}\right)^2$$

You have made some errors which prevent a nice compact end result. :D

 

[Dethrone]

Oopps, I mistyped a line. Should have doubled checked, but still something wrong.

$$V=\pi \int_{0}^{b\sqrt{a}} a^2y-\frac{2a}{3b}y^3+\frac{y^5}{5b}\,dy$$

should be:

$$V=\pi (a^2y-\frac{2a}{3b}y^3+\frac{y^5}{5b})\,|_{0}^{b\sqrt{a}}$$

which I got from integrating this:

$$V=\pi \int_{0}^{b\sqrt{a}} a^2-2a\frac{y^2}{b^2}+\frac{y^4}{b^4}\,dy$$

 

[MarkFL]

Okay, this is correct:

$$V=\pi\int_0^{b\sqrt{a}} a^2-\frac{2a}{b^2}y^2+\frac{1}{b^4}y^4\,dy$$

Now, try applying the FTOC...what is the anti-derivative?

 

[Dethrone]

Okay, this is correct:

$$V=\pi\int_0^{b\sqrt{a}} a^2-\frac{2a}{b^2}y^2+\frac{1}{b^4}y^4\,dy$$

Now, try applying the FTOC...what is the anti-derivative?

I think I'm getting these wrong because I was in a rush. For the other method, my $x$ transformed into an $a$ halfway through the problem so I ended up integrating the wrong variable. I checked this like 5 times, so it escaped me 5 times...(Tmi)

$$V=\pi (a^2y-\frac{2a}{b^2}\frac{y^3}{3}+\frac{1}{b^4}\frac{y^5}{5})|_0^{b\sqrt{a}}$$

If this is wrong, I will immediately enroll myself back in an algebra 1 class...(Speechless)

 

[MarkFL]

You will not have to enroll in an algebra class, that is correct...now finish out the FTOC...what do you get?

 

[Dethrone]

Went back and fixed my cylindrical method too, and voila! I got for both:

$$V=\frac{8}{15}\pi a^{5/2}b$$

Thanks Mark, as always! (Cool)

 

[MarkFL]

Yes, good work! (Yes)

 

[MarkFL]

By the way, you are doing yourself good by trying to use two methods on solids of revolution...not only is it a way to check your work, but it is good practice as well. :D

Sometimes it is impractical to use two methods, but when you can, I highly recommend it to all students.

 

[Dethrone]

Now for part two of the problem: (Nod)

Part II:
Suppose that $a$ changes with time and the axis moves horizontally at a rate of $\left(b\sqrt{a}-\frac{b^3}{\sqrt{a}}\right)/\tau$ m/s, where $\tau$ has units $$ and is a constant. Find the rate of increase of $V$ with time.

In part a), we determined that $V=\frac{8}{15}\pi a^{5/2}b$. Since $a$ will be moving horizontally with time, $b$ is still constant. Differentiating both sides w.r.t time, $t$:

$$\d{V}{t}=\frac{8}{15}\pi \frac{5}{2}a^{3/2}\d{a}{t}b$$

We know that $\d{a}{t}=\left(b\sqrt{a}-\frac{b^3}{\sqrt{a}}\right)/\tau$:$$\d{V}{t}=\frac{8}{15}\pi \frac{5}{2}a^{3/2}b\left(b\sqrt{a}-\frac{b^3}{\sqrt{a}}\right)/\tau $$
$$=\frac{4}{3}\frac{ab^2\left(a-b^2\right)}{\tau}$$

Is that correct?

Spoiler

There will be a Part III (Headbang)

 

[MarkFL]

Yep, looks good to me. (Yes)

 

[Dethrone]

Part III is just optimizing the volume, actually. I'm sure I can figure that out. Thanks Mark! :D