Surface of revolution - y = (1/3)x^3

[Dethrone]

Find the area of the surface of revolution $y=\frac{1}{3}x^3$ from $x=0$ to $3$, about the y-axis.

$$S=\int_{0}^{9} 2 \pi x \sqrt{1+(\d{x}{y}})^2\,dy$$
$$=2\pi\int_{0}^{9} x \sqrt{1+\frac{1}{x^4}}\,dy$$
$$=2\pi\int_{0}^{9} \frac{1}{x}\sqrt{x^4+1}\,dy$$
$$=2\pi\int_{0}^{9} \frac{1}{(3y)^{1/3}}\sqrt{(3y)^{4/3}+1}\,dy$$

At this point, I can only say, "Why couldn't $(3y)^{1/3}$ have been in the numerator!"... (Crying)(Crying)

This integral is quite hard to solve, and I was wondering if I missed something in the question. Now I did solve the integral, but not without a series of substitutions: let $3y = u$,let $u = z^3$, let $z^2 = a$, let $a = tan \theta$. Surely, this question could have be solved much easier?

EDIT: Not to mention, after the tangent substitution at the end, you get the dreaded $\sec^3\left({\theta}\right)$ requiring repeated integration by parts...(Swearing)

 

[MarkFL]

I think I would integrate with respect to $x$ to begin with:

$$S=2\pi\int_0^3 x\sqrt{x^4+1}\,dx$$

or:

$$S=2\pi\int_0^3 \frac{x^5+x}{\sqrt{x^4+1}}\,dx$$

or:

$$S=\frac{2\pi}{3}\int_0^3 \frac{2x^5+2x+x^5+x}{\sqrt{x^4+1}}\,dx$$

Now, can you split this integrand into 3 terms with well-known anti-derivatives?

 

[Dethrone]

Before I split the integrand, how come you have three different integrals? Are they all the same thing?

 

[MarkFL]

Before I split the integrand, how come you have three different integrals? Are they all the same thing?

Yes, in the first step I multiplied the integrand by:

$$1=\frac{\sqrt{x^4+1}}{\sqrt{x^4+1}}$$

and in the second step I am using:

$$x^5+x=\frac{3x^5+3x}{3}=\frac{2x^5+2x+x^5+x}{3}$$

 

[Dethrone]

I'm not sure if I'm seeing the anti-derivative you're looking for, but $\frac{1}{\sqrt{x^4+1}}$ reminds me of a hyperbolic sine antiderivative.

 

[MarkFL]

I'm not sure if I'm seeing the anti-derivative you're looking for, but $\frac{1}{\sqrt{x^4+1}}$ reminds me of a hyperbolic sine antiderivative.

Yes, what is:

$$\frac{d}{dx}\left(\sinh^{-2}\left(x^2\right)\right)$$ ?

 

[Dethrone]

That's $\frac{2x}{\sqrt{x^4+1}}$. I was thinking about making a substitution $x^2=z$, but I guess that's not necessary.

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So far:

$$\frac{2x}{\sqrt{x^4+1}} +\frac{2x^5+x^5+x}{\sqrt{x^4+1}}$$

I still need to split it once more.

 

[Dethrone]

It's much more evident to me when I'm using the substitution $x^2=z$.

$$=\int \frac{3x^5+3x}{\sqrt{x^4+1}} dx$$
$$=\int \frac{3z^{5/2}+3z^{1/2}}{z^2+1} \frac{dz}{2\sqrt{z}}$$
$$=3 \int \frac{z^2+1}{\sqrt{z^2+1}}dz$$
$$=3 (\int \sqrt{z^2+1})$$

This is how I would normally do it. I was a bit slow earlier because I was working from my laptop.

 

[MarkFL]

That's $\frac{2x}{\sqrt{x^4+1}}$.

Good! Yes, now we may write:

$$S=\frac{2\pi}{3}\int_0^3 \frac{2x}{\sqrt{x^4+1}}+\frac{2x^5+x^5+x}{\sqrt{x^4+1}}\,dx$$

or:

$$S=\frac{2\pi}{3}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\frac{1}{2}\int_0^3\frac{4x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx\right)$$

For the second integral, think "product rule" since we have a sum...

 

[Dethrone]

I'm not following your steps.

$$\int_{0}^{3} \frac{2x^5+x^5+x}{\sqrt{x^4+1}}\,dx = \frac{1}{2}\int_{0}^{3} \frac{4x^5}{\sqrt{x^4+1}}+\frac{2x^5+2x}{\sqrt{x^4+1}}\,dx$$

I can't seem to get this:

$$\frac{1}{2}\int_0^3\frac{4x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx$$

 

[MarkFL]

Factor the numerator in the second term, and then cancel...:D

 

[Dethrone]

To avoid excessive LaTeX, I will work on the second half of the integral.

$$\frac{1}{2}\int_0^3\frac{4x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx$$

By product rule, are you referring to f'(x)g(x) + g'(x)f(x), or are you talking about integration by parts?

 

[MarkFL]

I am referring to the product rule for differentiation. See if you can express that integrand as the product of a differentiation. :D

 

[Dethrone]

The only thing I see is

$$\d{}{x}[\sqrt{x^4+1}(x^2)]=\frac{4x^3}{2\sqrt{x^4+1}}+2x\sqrt{x^4+1}$$

But it's still not the integrand. Let me know if I'm going in the right direction, I'll try harder. This seems like a unique way to do integration, I don't think I've ever read about this.

 

[MarkFL]

The only thing I see is

$$\d{}{x}[\sqrt{x^4+1}(x^2)]=\frac{4x^3}{2\sqrt{x^4+1}}+2x\sqrt{x^4+1}$$

But it's still not the integrand. Let me know if I'm going in the right direction, I'll try harder. This seems like a unique way to do integration, I don't think I've ever read about this.

You are headed in the right direction...but your application of the product rule is a bit off...:D

 

[Dethrone]

Dear me, now I can't even differentiate. I've overexerted myself today...

So, I would write it like this? :P

$$\int d\left(x^2(\sqrt{x^4+1})\right)$$

Can you explain this notation a bit? I've never seen it before. By integrating with respect to the function, you are essentially cancelling out the integration?

 

[MarkFL]

I can see I am a bit off here...let's return to:

$$S=\frac{2\pi}{3}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\frac{2x^5}{\sqrt{x^4+1}}+x\sqrt{x^4+1}\,dx\right)$$

Now, let's write this as:

$$S=\frac{2\pi}{3}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\frac{2x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}-x\sqrt{x^4+1}\,dx\right)$$

or:

$$S=\frac{2\pi}{3}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\frac{2x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx\right)-\frac{1}{3}S$$

Hence:

$$\frac{4}{3}S=\frac{2\pi}{3}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\frac{2x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx\right)$$

or:

$$S=\frac{\pi}{2}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\frac{2x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx\right)$$

Okay, now we can apply the product rule I was talking about before...:D

 

[MarkFL]

Dear me, now I can't even differentiate. I've overexerted myself today...

So, I would write it like this? :P

$$\int d\left(x^2(\sqrt{x^4+1})\right)$$

Can you explain this notation a bit? I've never seen it before. By integrating with respect to the function, you are essentially cancelling out the integration?

Yes, that's a good way to think of it:

$$\int\,du=u+C$$

So, whatever $u$ is, the integral of its differential is just $u$ itself...the two operations are inverses of one another and so the indeed do "cancel" out (plus the constant of integration).

 

[Dethrone]

Oh. My. God. How does one LaTeX so quickly...(Speechless)

$$\displaystyle S=\frac{\pi}{2}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\,d\left((x^2)(\sqrt{x^4+1}\right)\right)$$

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Also, how did you plan out and see the product rule...? I would have never seen that as the product rule...

 

[MarkFL]

Oh. My. God. How does one LaTeX so quickly...(Speechless)

$$\displaystyle S=\frac{\pi}{2}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\,d\left((x^2)(\sqrt{x^4+1}\right)\right)$$

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Also, how did you plan out and see the product rule...? I would have never seen that as the product rule...

I usually copy/paste my last line, and then modify it as needed, then repeat. :D

I cheated, looked at the anti-derivative as given by W|A, and then worked backwards in an effort to look slick. (Cool)

 

[Dethrone]

Aww, I thought it was a trick I could learn :(
But it was very clever indeed. Now getting back to the original question, how would differentiating with respect to x first help with finding the answer?

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I also realized that you avoided a variable substitution. Was there a reason for that, or just personal preference?

This is what I would have normally done:$$=\int \frac{3x^5+3x}{\sqrt{x^4+1}} dx$$
$$=\int \frac{3z^{5/2}+3z^{1/2}}{\sqrt{z^2+1}} \frac{dz}{2\sqrt{z}}$$
$$=3 \int \frac{z^2+1}{\sqrt{z^2+1}}dz$$
$$=3 \int \sqrt{z^2+1} dz$$
$$=3 \int \sec^3\left({\theta}\right) d\theta$$

 

[MarkFL]

Aww, I thought it was a trick I could learn :(
But it was very clever indeed. Now getting back to the original question, how would differentiating with respect to x first help with finding the answer?

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I also realized that you avoided a variable substitution. Was there a reason for that, or just personal preference?

This is what I would have normally done:$$=\int \frac{3x^5+3x}{\sqrt{x^4+1}} dx$$
$$=\int \frac{3z^{5/2}+3z^{1/2}}{\sqrt{z^2+1}} \frac{dz}{2\sqrt{z}}$$
$$=3 \int \frac{z^2+1}{\sqrt{z^2+1}}dz$$
$$=3 \int \sqrt{z^2+1} dz$$
$$=3 \int \sec^3\left({\theta}\right) d\theta$$

Well, there actually is a trick to be learned...that rewriting the integrand and looking for known derivatives can be useful. :D

Integrating the cube of the secant function can be fun too...

With a bit of algebra, we may write this in a form we may integrate directly:

$\displaystyle \sec^3(\alpha)=\frac{1}{2}(\sec^3(\alpha)+\sec^3(\alpha))=\frac{1}{2}(\sec^3(\alpha)+\sec(\alpha)(\tan^2(\alpha)+1))=$

$\displaystyle \frac{1}{2}\left(\sec^3(\alpha)+\sec(\alpha)\tan^2(\alpha)+\sec(\alpha)\frac{\tan(\alpha)+\sec(\alpha)}{\tan(\alpha)+\sec(\alpha)} \right)=$

$\displaystyle \frac{1}{2}\left(\sec(\alpha)\sec^2(\alpha)+\sec(\alpha)\tan^2(\alpha)+\frac{\sec(\alpha)\tan(\alpha)+\sec^2(\alpha)}{\sec(\alpha)+\tan(\alpha)} \right)=$

$\displaystyle \frac{1}{2}\frac{d}{d\theta}\left(\sec(\alpha)\tan(\alpha)+\ln\left|\sec(\alpha)+\tan(\alpha) \right| \right)$

 

[Dethrone]

While looking for "derivatives" is a good trick, if there aren't any, wouldn't you be wasting more time than if you were to use a more standard technique? For example, the "product rule" trick only worked because of W|A.

About $\sec^3\left({x}\right)$, your method seems much more fun than using integration my parts...

Also, getting back to the original question:

$$\displaystyle S=\frac{\pi}{2}\left(\int_0^3 d\left(\sinh^{-1}\left(x^2\right)\right)+\int_0^3\frac{2x^5}{\sqrt{x^4+1}}+2x\sqrt{x^4+1}\,dx\right)$$
$$=\frac{\pi}{2}\ln\left({x^2+\sqrt{x^4+1}}\right)|_0^3 + \frac{\pi}{2}\left(x^2\sqrt{x^4+1}\right)|_0^3 $$

This gives the right answer, but why are we allowed to differentiate w.r.t x when we're rotating about the y-axis?

 

[MarkFL]

I would say that if a standard technique yields results, then take that path, but sometimes we may wander a bit before finding a path that leads to the fruits of a result.

I simply began with the formula in my old Calc textbook:

$$S=2\pi\int_a^b x\sqrt{1+\left(f'(x)\right)^2}\,dx$$

The $x$ in the integrand represents the radius of the element of the surface, while the radical and differential are the width of the element.

 

[Dethrone]

Take makes sense. The book I'm using makes it clear that rotation about the y-axis requires
$$S=\int 2 \pi x \sqrt{1+\left(\d{x}{y}\right)^2}\,dy$$

and rotation about the x-axis requires:

$$S=\int 2 \pi y \sqrt{1+\left(\d{y}{x}\right)^2}\,dx$$

But it's not always the case. Right now, it seems that the only thing that is constant for rotation about the y-axis is that the radius must be "x" which makes sense, but the width can either be differentiated with respect to x or y as long as you integrate with respect to that too.

Thanks for the help, it makes sense now! It's about 3am here, so I'm going to bed (Yawn)

 

[MarkFL]

We're in the same time zone then...and I am feeling the droopiness beginning to invade my eyes as well. :D

 

[Dethrone]

Now that I reread this question, what was your original motive for splitting up the integrand like this?

$$\displaystyle S=\frac{2\pi}{3}\int_0^3 \frac{2x^5+2x+x^5+x}{\sqrt{x^4+1}}\,dx$$

You said it would be split into 3 integrals, but it only became two in the end.

 

[MarkFL]

Now that I reread this question, what was your original motive for splitting up the integrand like this?

$$\displaystyle S=\frac{2\pi}{3}\int_0^3 \frac{2x^5+2x+x^5+x}{\sqrt{x^4+1}}\,dx$$

You said it would be split into 3 integrals, but it only became two in the end.

One of the three resulting integrals was a constant times $S$, so it was moved to the other side of the equation. :D

 

[MarkFL]

Find the area of the surface of revolution $y=\frac{1}{3}x^3$ from $x=0$ to $3$, about the y-axis.

$$S=\int_{0}^{9} 2 \pi x \sqrt{1+(\d{x}{y}})^2\,dy$$
$$=2\pi\int_{0}^{9} x \sqrt{1+\frac{1}{x^4}}\,dy$$
$$=2\pi\int_{0}^{9} \frac{1}{x}\sqrt{x^4+1}\,dy$$
$$=2\pi\int_{0}^{9} \frac{1}{(3y)^{1/3}}\sqrt{(3y)^{4/3}+1}\,dy$$

At this point, I can only say, "Why couldn't $(3y)^{1/3}$ have been in the numerator!"... (Crying)(Crying)

This integral is quite hard to solve, and I was wondering if I missed something in the question. Now I did solve the integral, but not without a series of substitutions: let $3y = u$,let $u = z^3$, let $z^2 = a$, let $a = tan \theta$. Surely, this question could have be solved much easier?

EDIT: Not to mention, after the tangent substitution at the end, you get the dreaded $\sec^3\left({\theta}\right)$ requiring repeated integration by parts...(Swearing)

It bothered me that I provided help by using a CAS to get the result and then work backwards, so let's have another go at this.

Using the formula:

$$S=2\pi\int_a^b x\sqrt{1+\left(\d{y}{x}\right)^2}\,dx$$

we identify:

$$y=\frac{1}{3}x^3\,\therefore\,\d{y}{x}=x^2$$

And we have:

$$S=2\pi\int_0^3 x\sqrt{\left(x^2\right)^2+1}\,dx$$

Let:

$$u=x^2\,\therefore\,du=2x\,dx$$

And we have:

$$S=\pi\int_0^9 \sqrt{u^2+1}\,du$$

Now, using the substitution:

$$u=\tan(\theta)\,\therefore\,du=\sec^2(\theta)\,d\theta$$

We get:

$$S=\pi\int_0^{\tan^{-1}(9)} \sec^3(\theta)\,d\theta$$

Using the formula I derived earlier:

$$\sec^3(\theta)=\frac{1}{2}\frac{d}{d\theta}\left(\sec(\theta)\tan(\theta)+\ln\left|\sec(\theta)+\tan(\theta) \right| \right)$$

We obtain:

$$S=\frac{\pi}{2}\int_0^{\tan^{-1}(9)}\,d\left(\sec(\theta)\tan(\theta)+\ln\left|\sec(\theta)+\tan(\theta) \right| \right)$$

Application of the FTOC yields:

$$S=\frac{\pi}{2}\left[\sec(\theta)\tan(\theta)+\ln\left|\sec(\theta)+\tan(\theta) \right|\right]_0^{\tan^{-1}(9)}$$

Given:

$$\theta=\tan^{-1}(z)\implies\sec(\theta)=\sqrt{z^2+1}$$

there results:

$$S=\frac{\pi}{2}\left(\left(9\sqrt{82}+\ln\left(\sqrt{82}+9\right)\right)-\left(0\right)\right)$$

$$S=\frac{\pi}{2}\left(9\sqrt{82}+\ln\left(\sqrt{82}+9\right)\right)$$

 

[Dethrone]

Thanks Mark!

Spoiler

Thanks for working it out again. I was beginning to doubt my ability to integrate yesterday when I failed to see where you were going with your pointers - since I failed to notice the backwards application of the product rule. Though, I will keep that trick in mind, and it's also not too difficult to notice it. :D

Seems that $\int \sec^3\left({x}\right)$ is mostly unavoidable unless you're a CAS, but I really like how you rewrote $\sec^3\left({x}\right)$ in a form that is easy to integrate. For some reason, I hate integration by parts, and I really don't find it useful since the need to use it rarely comes up in the problems I do.

At the end of your solving, did you end up plugging $\tan^{-1}(9)$ in? Or did you convert the bounds to $z$? I've always avoided changing bounds to that of radians because of situations like this.