Find the volume bounded by hyperboloid and plane z = ± d

[songoku]

Homework Statement

Please see below

Relevant Equations

Double Integral
Triple Integral
Cylindrical coordinate
Spherical coordinate

My attempt:
The shape of the hyperboloid would be like this:

If the hyperbolod is cut by plane z = d, the intersection would be a ellipse. Projecting the intersection to xy - plane, I think I get:
$$-2\leq x \leq 2$$
$$-b\sqrt{1-\frac{x^2}{a^2}} \leq y \leq b\sqrt{1-\frac{x^2}{a^2}}$$

So the volume would be:
$$V=2 \times \int_{-2}^{2} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \left(c\sqrt{\frac{x^2}{a^2}+\frac{y^2}{b^2}-1}\right)dydx$$

Is this correct?

Thanks

 

[Orodruin]

Projecting the intersection to xy - plane, I think I get:
$$-2\leq x \leq 2$$

Where did the 2 come from? This seems entirely random to me.

$$-b\sqrt{1-\frac{x^2}{a^2}} \leq y \leq b\sqrt{1-\frac{x^2}{a^2}}$$

For example, what happens here if $a = 0.1$ and $x=2$?

So the volume would be:
$$V=2 \times \int_{-2}^{2} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \left(c\sqrt{\frac{x^2}{a^2}+\frac{y^2}{b^2}-1}\right)dydx$$

This cannot be right. For example, think about what happens to the integrand when $x=y=0$.

Is this correct?

No. I would suggest a different approach entirely. You seem to want to integrate the z value over the relevant x-y region but this is unnecessarily difficult. For example, inside the ellipse $(x/a)^2 + (y/b)^2=1$, the height of the desired region is $2d$. Outside of that region it is given by a different expession.

 

[pasmith]

I would use scaled cylindrical coordinates $(r, \theta, \zeta)$, where $$\begin{split} x &= ar\cos \theta, \\ y &= br\sin \theta, \\ z &= c\zeta. \end{split}$$

 

[nuuskur]

Where do you get $-2\leqslant x\leqslant 2$?

Putting $z=0$ we see that the shape is an ellipse. So we are stacking "elliptical cylinders" on top of each other. Figure out how the volume of each such cylinder changes based on $0\leqslant z\leqslant d$ and then integrate.

Fix $z$ and note that
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{z^2}{c^2} \Leftrightarrow \frac{x^2}{a(z)^2} + \frac{y^2}{b(z)^2} = 1,$$
where
$$a(z)^2 = a^2 \frac{c^2+z^2}{c^2}\quad\mbox{and}\quad b(z)^2 = b^2\frac{c^2+z^2}{c^2}.$$
Area of ellipse is $\pi ab$.

 

[Orodruin]

I would use scaled cylindrical coordinates $(r, \theta, \zeta)$, where $$\begin{split} x &= ar\cos \theta, \\ y &= br\sin \theta, \\ z &= c\zeta. \end{split}$$

I would just sum up the volumes of infinitesimally thick elliptical discs … aka slicing method

 

[songoku]

I would use scaled cylindrical coordinates $(r, \theta, \zeta)$, where $$\begin{split} x &= ar\cos \theta, \\ y &= br\sin \theta, \\ z &= c\zeta. \end{split}$$

Sorry I don't think I have covered scaled cylindrical coordinates. Should it be taught in the same part as cylindrical coordinates or in different part?

Where did the 2 come from? This seems entirely random to me.

Where do you get $-2\leqslant x\leqslant 2$?

My bad. It should be: $-a\leq x \leq a$

For example, what happens here if $a = 0.1$ and $x=2$?This cannot be right. For example, think about what happens to the integrand when $x=y=0$.

It would be imaginary.

You seem to want to integrate the z value over the relevant x-y region but this is unnecessarily difficult. For example, inside the ellipse $(x/a)^2 + (y/b)^2=1$, the height of the desired region is $2d$. Outside of that region it is given by a different expession.

Yes you are correct, that was my idea

Putting $z=0$ we see that the shape is an ellipse. So we are stacking "elliptical cylinders" on top of each other. Figure out how the volume of each such cylinder changes based on $0\leqslant z\leqslant d$ and then integrate.

I would just sum up the volumes of infinitesimally thick elliptical discs … aka slicing method

I can imagine the slicing and stacking part, so based on this hint I am imagining the integral would be something like this:
$$V=2\times \int_{0}^{d}f(z) dz$$

where $f(z)$ is the expression of the slice of ellipse being stacked upon. But I am struggling to find the appropriate expression for $f(z)$

Fix $z$ and note that
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{z^2}{c^2} \Leftrightarrow \frac{x^2}{a(z)^2} + \frac{y^2}{b(z)^2} = 1,$$
where
$$a(z)^2 = a^2 \frac{c^2+z^2}{c^2}\quad\mbox{and}\quad b(z)^2 = b^2\frac{c^2+z^2}{c^2}.$$
Area of ellipse is $\pi ab$.

I can follow the algebra but sorry I don't understand how to apply this to the question. Should I make z the subject of the equation (since I want to find $f(z)$)?

Thanks

 

[Orodruin]

What is the volume of a cylinder with height $dz$ and base area $A(z)$?

 

[songoku]

What is the volume of a cylinder with height $dz$ and base area $A(z)$?

$A(z) dz$

 

[nuuskur]

My bad. It should be: $-a\leq x \leq a$

This cannot be correct. The permitted values of $x$ change based on $z$ - consider the graph. $a$ is just some fixed number, it does not depend on $z$.

It would be imaginary.

Stay within $\mathbb R^3$. There is no reason to overcomplicate.

I can imagine the slicing and stacking part, so based on this hint I am imagining the integral would be something like this:
$$V=2\times \int_{0}^{d}f(z) dz$$

Yes!

where $f(z)$ is the expression of the slice of ellipse being stacked upon. But I am struggling to find the appropriate expression for $f(z)$

It is the area of the ellipse for a fixed value of $z$.

 

[Orodruin]

This cannot be correct. The permitted values of x change based on z - consider the graph. a is just some fixed number, it does not depend on z.

Consider what the OP was trying to do here. They were trying to write an xy-integral of the height of the object. As such, the dependent variable would be z and not x or y. The conclusion was still incorrect, but the x-range cannot depend on z because in OP’s approach it is the other way around. (The x-range does however depend on $d$)

Stay within R3. There is no reason to overcomplicate.

That comment from the OP was a reply to my comment regarding what happened to the OP’s expression in a particular case. In that case, yes, the integrand became imaginary so it was a correct conclusion regarding the integrand. The following conclusion should be that the expression could not be correct as the volume is not complex.

$A(z) dz$

Indeed, so when you sum up those volumes you obtain the full volume
$$
V = \int_{-d}^d A(z) dz.
$$
What is the function $A(z)$ expressed as?

 

[pasmith]

I would just sum up the volumes of infinitesimally thick elliptical discs … aka slicing method

But how do you calculate the area of an ellipse?

 

[Orodruin]

But how do you calculate the area of an ellipse?

Scaling the expression for the area of a circle of unit radius

 

[songoku]

It is the area of the ellipse for a fixed value of $z$.

Scaling the expression for the area of a circle of unit radius

Taking a fix value of $z$ to be $k$, the equation becomes:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1+\frac{k^2}{c^2}$$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{c^2+k^2}{c^2}$$
$$\frac{x^2}{\frac{a^2(c^2+k^2)}{c^2}}+\frac{y^2}{\frac{b^2(c^2+k^2)}{c^2}}=1$$
$$\frac{x^2}{\left(\frac{a}{c}\sqrt{c^2+k^2}\right)^2}+\frac{y^2}{\left(\frac{b}{c}\sqrt{c^2+k^2}\right)^2}$$

The area is $\pi \left(\frac{a}{c}\sqrt{c^2+k^2}\right)\left(\frac{b}{c}\sqrt{c^2+k^2}\right)=\pi\frac{ab}{c^2}(c^2+k^2)$

So my integrand maybe would be: $\pi\frac{ab}{c^2}(c^2+z^2)$

Is this correct? Thanks

 

[nuuskur]

Missing an $..=1$ for the last centered expression. Otherwise, yes, $2\int _0^d A(z)dz$, where $A(z)$ is the quantity you calculated, is a correct way to obtain the volume.

 

[songoku]

Thank you very much for all the help and explanation Orodruin, pasmith, nuuskur