Finding Volume Using Double Integrals: A Question on Cylindrical Coordinates

[peripatein]

Hi,

Homework Statement

I am asked to find the volume under the curve whose equation is z=16-(x^4+y^4), and within (x^2+y^2)<=1, using a double integral.

Homework Equations

The Attempt at a Solution

Should I use cylindrical coordinates?
I feel slightly lost. I have tried drawing z=16-(x^4+y^4), unsuccessfully.
I understand that (x^2+y^2)<=1 is a circle of r<=1 and sets the limit to the volume, i.e. the base of the shape.
I'd truly appreciate some guidance.

 

[DryRun]

Here's the plot for $z=16-(x^4+y^4)$:
http://img8.uploadhouse.com/fileuploads/17053/17053388c9a8698ea1f3d43b18b267345d128d42.gif
Cylindrical coordinates should work fine. First, convert the equations of the curve and circle into polar coordinates.

 

[Dick]

Hi,

Homework Statement

I am asked to find the volume under the curve whose equation is z=16-(x^4+y^4), and within (x^2+y^2)<=1, using a double integral.

Homework Equations

The Attempt at a Solution

Should I use cylindrical coordinates?
I feel slightly lost. I have tried drawing z=16-(x^4+y^4), unsuccessfully.
I understand that (x^2+y^2)<=1 is a circle of r<=1 and sets the limit to the volume, i.e. the base of the shape.
I'd truly appreciate some guidance.

These are surfaces, not curves. It's sort of a boxy 'paraboloid like thing' opening down. You really don't need to know much about the detailed shape. The important thing about it is that it lies completely above the disk x^2+y^2<=1 in the xy plane. Having said that I now realize I was assuming the xy plane was the lower bound. x^2+y^2<=1 is actually the inside of an infinite cylinder. If there is not some other z cutoff the volume will be infinite. Is that exactly how the problem is stated?

 

[DryRun]

Having said that I now realize I was assuming the xy plane was the lower bound. x^2+y^2<=1 is actually the inside of an infinite cylinder. If there is not some other z cutoff the volume will be infinite. Is that exactly how the problem is stated?

That's exactly what i thought, but then there would be no purpose to calculating the volume, as the answer would be directly obtained theoretically. I believe it's a wrongly worded problem and the plane $z=0$ should be the lower limit. The OP should have provided the complete question accurately.

 

[peripatein]

The volume bounded under the tent-like shape given by the equation z=16-(x^4+y^4), and within the area (x^2+y^2)<=1.
Is it clearer now?

 

[Dick]

The volume bounded under the tent-like shape given by the equation z=16-(x^4+y^4), and within the area (x^2+y^2)<=1.
Is it clearer now?

z=16-(x^4+y^4) defines the top, the cylinder x^2+y^2=1 defines the sides and the bottom is defined by z=0 i.e the xy plane. Is that what you mean? The 'bottom' needs to somehow define the lowest value of z.

 

[peripatein]

So z changes from 0 to 16-(x^4+y^4) while y changes from 0 to sqrt(1-x^2) (the volume then to be multiplied by two)?

 

[SammyS]

So z changes from 0 to 16-(x^4+y^4) while y changes from 0 to sqrt(1-x^2) (the volume then to be multiplied by two)?

@peripatein,

All the respondents are asking you, "Is the region bounded by z=0 below?" You're the one with access to the problem that you're trying to solve.

 

[peripatein]

But I have already stated the problem exactly as it is presented to me, verbatim! I am in possession of no further information.

 

[DryRun]

But I have already stated the problem exactly as it is presented to me, verbatim! I am in possession of no further information.

In that case, you should inform your teacher about the missing lower bound for z. If it's a problem you got from a book, it is missing some required information.

Anyway, when you get the equation of the lower plane (we had assumed it was $z=0$), just use that value as the lower limit of $z$ in your double integral.

 

[peripatein]

May I, in that case, say that z changes from 0 to 16-(x^4+y^4) while y changes from 0 to sqrt(1-x^2) (in which case the volume would then have to be multiplied by 2)?

 

[DryRun]

May I, in that case, say that z changes from 0 to 16-(x^4+y^4) while y changes from 0 to sqrt(1-x^2) (in which case the volume would then have to be multiplied by 2)?

If you are evaluating the volume using Cartesian coordinates, you need to specify the limits for $x$ as well.

Using Cartesian coordinates, if you say that x varies from 0 to $\sqrt (1-y^2)$, the volume obtained after evaluating the triple integral would need to be multiplied by 4, since you would then have calculated the volume of the solid in the first octant only. But if you state that x varies from $-\sqrt (1-y^2)$ to $\sqrt (1-y^2)$, then in that case, you would have to multiply by 2, in order to get the final volume.

But it would be easier to use cylindrical coordinates. Refer to post #2.

 

[peripatein]

As I am requested to use a double integral, using cylindrical coordinates should r change from 0 to 1 while z changes from 0 to 16-(x^4+y^4), with Jacobian of transformation r(dr)(dtheta)(dz)? Or do I need to let theta change from 0 to 2pi? I am not sure.

 

[DryRun]

As I am requested to use a double integral, using cylindrical coordinates should r change from 0 to 1 while z changes from 0 to 16-(x^4+y^4), with Jacobian of transformation r(dr)(dtheta)(dz)? Or do I need to let theta change from 0 to 2pi? I am not sure.

You need to provide the limits for the following 3 parameters: r, θ and z.
If θ varies from 0 to 2∏, you'll get a complete revolution about the z-axis, forming the entire volume of the solid.
You also need to re-write the equation of the curve in terms of polar coordinates before you proceed with evaluating the triple integral.

 

[SammyS]

As I am requested to use a double integral, using cylindrical coordinates should r change from 0 to 1 while z changes from 0 to 16-(x^4+y^4), with Jacobian of transformation r(dr)(dtheta)(dz)? Or do I need to let theta change from 0 to 2pi? I am not sure.

You are referring to 3 variables here, r, θ, and z.

To find the volume with a double integral rather than a triple integral, you need to have an integrand that's different than 1.

[STRIKE]The obvious integrand, in my opinion, is 16-r⁴, and integrate over r and θ, using the appropriate Jacobian.

Alternatively, you could use an integrand of 2π r and integrate over r and z.[/STRIKE]

Added in Edit:
I've crossed out the baloney !

 

[peripatein]

Is z not equal to 16-r^4(cos(2theta))^2, where r is <= 1?

 

[DryRun]

$x^4+y^4=r^4$ is easily obtained from the regular expression $x^2+y^2=r^2$, if you substitute $x=x^2$ and $y=y^2$ into the previous formula.

Since $z=16-(x^4+y^4)$ is now $z=16-r^4$, at $r=1$, what is the value of $z$? To find the maximum value of $z$, set $r=0$, which is the origin.

 

[peripatein]

It is still not very clear to me. First of all, why is that substitution (namely, x=x^2, y=y^2) even valid? Second, if z changes from 15 to 16, do I substitute these numerical values or do I use parameters still? What should be substituted for the (dz) element of the Jacobian? (-4r^3)dr?

 

[CAF123]

$x^4+y^4=r^4$ is easily obtained from the regular expression $x^2+y^2=r^2$, if you substitute $x=x^2$ and $y=y^2$ into the previous formula.

I am a little confused here too. If $$(x^2 + y^2) = r^2,\, \text{then}\, (x^2 + y^2)^2 = (r^2)^2.$$ This implies $$r^4 = (x^2 + y^2)^2 \neq x^4 + y^4?$$

 

[CAF123]

So you are effectively saying that $cos^4x + sin^4x = 1?$ While this holds for $x=0$, it does not for $x = \pi/4$.

http://www.wolframalpha.com/input/?i=cos^4x+++sin^4x

EDIT: Just so anyone else who reads this does not get confused, this was in response to a post that has now been deleted.

 

[DryRun]

I was mistaken in my previous attempt to explain the validity of $x^4+y^4=r^4$. This should be correct, i think.

From, $x^2+y^2=r^2$, let $x=a^2$, let $y=b^2$ and let $r=c^2$, then according to the formula, $(a^2)^2+(b^2)^2=(c^2)^2$.

Simplifying, gives: $a^4+b^4=c^4$. This is based on the Pythagoras' theorem. In a right-angled triangle, if the two sides that meet at a right angle are $a^2$ and $b^2$, then the hypotenuse must be: $(a^4 + b^4)$.

 

[CAF123]

I was mistaken in my previous attempt to explain the validity of $x^4+y^4=r^4$. This should be correct, i think.

From, $x^2+y^2=r^2$, let $x=a^2$, let $y=b^2$ and let $r=c^2$, then according to the formula, $(a^2)^2+(b^2)^2=(c^2)^2$.

Simplifying, gives: $a^4+b^4=c^4$. This is based on the Pythagoras' theorem.

I think this is correct now, because then we have $$|x|^4 + |y|^4 = r^4$$ and $|x|^4 = x^4$ etc..

How does this conform with what I wrote in post #19?

 

[DryRun]

How does this conform with what I wrote in post #19?

What you've done in #19 is simple algebraic manipulation by squaring both sides. But the problem in this particular case (Cartesian to polar coordinates) requires the use of the Pythagoras' theorem, since the basic expression: $x^2 + y^2 = r^2$ is obtained from that theorem.

It is still not very clear to me. First of all, why is that substitution (namely, x=x^2, y=y^2) even valid? Second, if z changes from 15 to 16, do I substitute these numerical values or do I use parameters still? What should be substituted for the (dz) element of the Jacobian? (-4r^3)dr?

Your first question should have been answered in the replies above.

You only need to set the limits of z. If you decide to put in every value of z as it varies, it defies the purpose of using integral to simplify the task. dz stays unchanged.

Here is your triple integral in terms of polar coordinates. You only have to set the limits.
$$\int \int \int rdrd\theta dz$$
It's a good idea to describe the region of integration to set the limits correctly:
For θ and r fixed, z varies from $z=0$ (as discussed previously, we have assumed that the lower bound is the plane z = 0) to $z=16-r^4$
For θ fixed, r varies from r=0 to r=1
θ varies from θ=0 to θ=2∏

 

[peripatein]

Forgive me for asking so many questions and/or for making it more difficult for you to clarify these issues, but I am still not sure I understand. Why did you include in your previous post just now a triple integral whereas I could only use a double?

 

[CAF123]

@peripatein
If you want to use a double integral then that's fine too. The two forms are equivalent.
By using the triple integral you get:$$ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{16-r^4} r\,dz\,dr\,d\theta$$
By doing the double integral you get: $$ \int_0^{2\pi} \int_0^1 16 - r^4\, r\,dr\,d\theta$$ Because you want a volume, it is implicitly assumed that the surface lies above z = 0, and it's projection will be something in the xy plane (i.e where z = 0)

 

[Dick]

I think this is correct now, because then we have $$|x|^4 + |y|^4 = r^4$$ and $|x|^4 = x^4$ etc..

How does this conform with what I wrote in post #19?

x^4+y^4 is NOT equal to r^4. The 'proof' is baloney. a^2+b^2=c^2 if a,b,c form the sides of a right triangle. That doesn't mean anything squared plus anything squared equals anything squared.

 

[pasmith]

$x^4+y^4=r^4$ is easily obtained from the regular expression $x^2+y^2=r^2$, if you substitute $x=x^2$ and $y=y^2$ into the previous formula.

No.
$$x^4 + y^4 = \frac12(x^2 + y^2)^2 + \frac12(x^2 - y^2)^2 = \frac12 r^4 + \frac12 r^4(\cos^2\theta - \sin^2\theta)^2 = \frac12 r^4 + \frac12 r^4(\cos(2\theta))^2 \\ = \frac12 r^4 (1 + \cos^2(2\theta)) \not\equiv r^4$$

Or, more directly, $x^4 + y^4 = r^4(\cos^4\theta + \sin^4\theta) \not\equiv r^4$ but the former expression is easier to integrate with respect to $\theta$.

 

[CAF123]

x^4+y^4 is NOT equal to r^4. The 'proof' is baloney. a^2+b^2=c^2 if a,b,c form the sides of a right triangle. That doesn't mean anything squared plus anything squared equals anything squared.

Hi Dick. I was initially confused with the statement $x^4 + y^4 = r^4$, however, where is the fault with sharks argument?

He let $x = a^2, y = b^2 and r = c^2$. Subbing this into $x^2 + y^2 = r^2$ gives $a^4 + b^4 = r^4$. But $a = |x|,$ so $a^4 = |x|^4 = x^4$ and similar reasoning for $b$.

 

[pasmith]

Hi Dick. I was initially confused with the statement $x^4 + y^4 = r^4$, however, where is the fault with sharks argument?

He let $x = a^2, y = b^2 and r = c^2$. Subbing this into $x^2 + y^2 = r^2$ gives $a^4 + b^4 = r^4$. But $a = |x|,$ so $a^4 = |x|^4 = x^4$ and similar reasoning for $b$.

The fact that $a^2 + b^2 = c^2$ does not imply that $a^4 + b^4 = c^4$: for example if $a = \sqrt 3$, $b = 2$ and $c = \sqrt 7$ then $a^2 + b^2 = c^2$, but $a^4 + b^4 = 25$ while $c^4 = 49$.

 

[SammyS]

You are referring to 3 variables here, r, θ, and z.

To find the volume with a double integral rather than a triple integral, you need to have an integrand that's different than 1.

The obvious integrand, in my opinion, is 16-r⁴, and integrate over r and θ, using the appropriate Jacobian.

Alternatively, you could use an integrand of 2π r and integrate over r and z.

Oh my goodness ! (Actually, not so good!)

It was me --- OK, Sheldon, I stand corrected. It was I --- who essentially said x⁴ + y⁴ = r⁴ .

Of course that's baloney ! (or is it bologna?) I've got to go to bed earlier.

There are several places in this thread where it would be helpful if responders would use the QUOTE feature when responding to a particular post.

 

[CAF123]

The fact that $a^2 + b^2 = c^2$ does not imply that $a^4 + b^4 = c^4$: for example if $a = \sqrt 3$, $b = 2$ and $c = \sqrt 7$ then $a^2 + b^2 = c^2$, but $a^4 + b^4 = 25$ while $c^4 = 49$.

I don't think we are using the fact that $a^2 + b^2 = c^2,$only that $x^2 + y^2 = r^2$ and making the substitutions $x = a^2, y = b^2, r = c^2$

 

[DryRun]

You are referring to 3 variables here, r, θ, and z.

To find the volume with a double integral rather than a triple integral, you need to have an integrand that's different than 1.

[STRIKE]The obvious integrand, in my opinion, is 16-r⁴, and integrate over r and θ, using the appropriate Jacobian.

Alternatively, you could use an integrand of 2π r and integrate over r and z.[/STRIKE]

Added in Edit:
I've crossed out the baloney !

I apologize for the mess, but honestly, i was unsure at first about the proof, but then a look at SammyS' (usually very trustworthy) post #15 stated that $16-r^4$, which meant that $x^4+y^4$ had to be equal to $r^4$. So, i tried to prove that statement.

Initially, i was going for the $\cos^4 θ +\sin^4 θ= 1$, but then i quickly realized it was wrong, as the expression didn't hold for all values of θ. Further effort seemed to indicate that the Pythagoras' theorem could be the right method. Thanks for crossing out the baloney!

No.
$$x^4 + y^4 = \frac12(x^2 + y^2)^2 + \frac12(x^2 - y^2)^2 = \frac12 r^4 + \frac12 r^4(\cos^2\theta - \sin^2\theta)^2 = \frac12 r^4 + \frac12 r^4(\cos(2\theta))^2 \\ = \frac12 r^4 (1 + \cos^2(2\theta)) \not\equiv r^4$$

Or, more directly, $x^4 + y^4 = r^4(\cos^4\theta + \sin^4\theta) \not\equiv r^4$ but the former expression is easier to integrate with respect to $\theta$.

Is there any other way of solving this?

 

[CAF123]

@Sharks Where is the fault in your argument?

 

[DryRun]

@Sharks Where is the fault in your argument?

Trying some values to verify that argument... Let the sides of a right-angled triangle be: a = 3, b = 4 and c (hypotenuse) = 5. This set of values works perfectly in the basic Pythagoras' theorem: $(3)^2+(4)^2=(5)^2$ giving $9+16=25$ which is correct.
Now, using that same set of values: $a^2=9, b^2=16$ and $c^2=25$. According to my previous argument, using the Pythagoras' theorem: $(9)^2+(16)^2$ should be equal to $(25)^2$ but, $81+256 \not = 625$. In theory it seemed like it would have worked, but i was clearly wrong.

 

[peripatein]

@peripatein
If you want to use a double integral then that's fine too. The two forms are equivalent.
By using the triple integral you get:$$ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{16-r^4} r\,dz\,dr\,d\theta$$
By doing the double integral you get: $$ \int_0^{2\pi} \int_0^1 16 - r^4\, r\,dr\,d\theta$$ Because you want a volume, it is implicitly assumed that the surface lies above z = 0, and it's projection will be something in the xy plane (i.e where z = 0)

Rather perplexing. Wishing to verify - may the integration, for obtaining the required volume, still be performed thus:
$$ \int_0^{2\pi} \int_0^1 16 - r^4\, r\,dr\,d\theta$$

?