Surface Intersection: Paraboloid & Plane

[evinda]

Hello! :D
Find the surface that is created by the intersection of the paraboloid $x^2+y^2-z=0$ and the plane $z=2$.

Is the intesection: $x^2+y^2=2,z=2$? Or is it not the asked surface?(Thinking)

 

[Chris L T521]

Hello! :D
Find the surface that is created by the intersection of the paraboloid $x^2+y^2-z=0$ and the plane $z=2$.

Is the intesection: $x^2+y^2=2,z=2$? Or is it not the asked surface?(Thinking)

I should first mention/clarify that whenever you take a plane and a surface in 3-space and find their intersection, the result is a curve, not a surface.

In order to find the curve that is generated when $x^2+y^2-z=0$ intersects $z=2$, all you need to do is substitute $z=2$ into the paraboloid equation as you did above to get $x^2+y^2=2$, which is a circle of radius $\sqrt{2}$.

Does this make sense? (Smile)

 

[HallsofIvy]

Hello! :D
Find the surface that is created by the intersection of the paraboloid $x^2+y^2-z=0$ and the plane $z=2$.

Is the intesection: $x^2+y^2=2,z=2$? Or is it not the asked surface?(Thinking)

Well, first, the intersection of two surface is, in general, NOT a surface, it is a curve! The intersection of these two surfaces, in particular, is the circle given by $x^2+ y^2= 2$ and $z= 2$.
That could be written as parametric equations $$x= \sqrt{2}cos(\theta)$$, $$y= \sqrt{2}sin(\theta)$$, $$z= 2$$.

 

[evinda]

So,isn't it possible to find a surface that is created by the intersection??
This: $x^2+y^2 \leq 2,z=2$ would be a surface,right?But this cannot be the asked surface,or am I wrong?? (Thinking)

 

[I like Serena]

So,isn't it possible to find a surface that is created by the intersection??
This: $x^2+y^2 \leq 2,z=2$ would be a surface,right?But this cannot be the asked surface,or am I wrong?? (Thinking)

Sure.
But since you've been asking questions before related to Green's, Stokes', and Gauss's theorems, I'd go one step further.

What you have there is the part of the surface that is a subset of the plane $z=2$, bounded by the intersection.
You have not identified the other part of that surface yet...

Together they form a surface that encapsulates a volume to which Gauss's theorem could be applied.

 

[evinda]

Sure.
But since you've been asking questions before related to Green's, Stokes', and Gauss's theorems, I'd go one step further.

What you have there is the part of the surface that is a subset of the plane $z=2$, bounded by the intersection.
You have not identified the other part of that surface yet...

Together they form a surface that encapsulates a volume to which Gauss's theorem could be applied.

I don't really know what to do...Could you give me a hint?? (Blush)

 

[I like Serena]

I don't really know what to do...Could you give me a hint?? (Blush)

Erm... that was the hint.

The other part of the surface is $x^2+y^2=2, z \le 2$.
That is the subset of the paraboloid bounded by the intersection.

 

[evinda]

Erm... that was the hint.

The other part of the surface is $x^2+y^2=2, z \le 2$.
That is the subset of the paraboloid bounded by the intersection.

So,these two surfaces: $x^2+y^2 \leq 2,z=2$ and $x^2+y^2=2, z \le 2$ are created by the intersection?? Could you explain me why?? (Thinking) (Blush)

 

[I like Serena]

So,these two surfaces: $x^2+y^2 \leq 2,z=2$ and $x^2+y^2=2, z \le 2$ are created by the intersection?? Could you explain me why??

The paraboloid looks like this:

View attachment 2117

The plane $z=2$ slices through it.
As a result you might visualize the bottom part of the cone sliced off, forming a bounded volume. The surface of this sliced off volume has 2 components: the paraboloid part at the bottom and the plane part at the top.

 

[evinda]

The paraboloid looks like this:

https://www.physicsforums.com/attachments/2117

The plane $z=2$ slices through it.
As a result you might visualize the bottom part of the cone sliced off, forming a bounded volume. The surface of this sliced off volume has 2 components: the paraboloid part at the bottom and the plane part at the top.

But.. are these $x^2+y^2 \leq 2$ and $z \leq 2$ formed by the intersection?? (Thinking) I haven't understood it... (Blush)

 

[I like Serena]

But.. are these $x^2+y^2 \leq 2$ and $z \leq 2$ formed by the intersection?? (Thinking) I haven't understood it... (Blush)

The actual intersection of those surfaces is the circle $x^2+y^2=2, z=2$.

Now compare the paraboloid to a cup.
We fill it with water up to the level z=2.

Then the surface of the water volume - all around - is the surface that is generated by the intersection.
We do not call this surface an intersection though.

 

[evinda]

The actual intersection of those surfaces is the circle $x^2+y^2=2, z=2$.

Now compare the paraboloid to a cup.
We fill it with water up to the level z=2.

Then the surface of the water volume - all around - is the surface that is generated by the intersection.
We do not call this surface an intersection though.

I understand...Thank you! :D

 

[evinda]

The actual intersection of those surfaces is the circle $x^2+y^2=2, z=2$.

Now compare the paraboloid to a cup.
We fill it with water up to the level z=2.

Then the surface of the water volume - all around - is the surface that is generated by the intersection.
We do not call this surface an intersection though.

I have now the solution of the exercise (Wait) I had to find the area of the surface.
It is done like that:

$$f(x,y,z)=x^2+y^2-z=0$$

$$\nabla{f}=(2x,2y,-1)$$

$$\nabla{f} \cdot \hat{k}=-1$$

$$\iint d \sigma= \iint \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}dA=\iint \frac{\sqrt{4x^2+4y^2+1}}{1}dxdy (****)$$

$$x=r \cos \theta , y=r \sin \theta$$

$$(****)=\int_{0}^{2 \pi} \int_0^{\sqrt{2}} \sqrt{ 4r^2+1}r dr d \theta= \dots= \frac{26 \pi }{6} $$But...what theorem have we used to find the area??

 

[I like Serena]

Hey again! (Happy)

I have now the solution of the exercise (Wait) I had to find the area of the surface.
It is done like that:

$$f(x,y,z)=x^2+y^2-z=0$$
$$\iint d \sigma= \iint \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}dA=\iint \frac{\sqrt{4x^2+4y^2+1}}{1}dxdy (****)$$

But...what theorem have we used to find the area??

They are using the formula:
$$\iint d \sigma= \iint \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}dA$$
where $d \sigma$ is a surface element in 3 dimensions and $dA$ is a surface element in the XY plane.
I do not believe it has a name.
If you want you can derive it, using that $\overrightarrow \nabla f$ is the (unnormalized) normal vector of the surface. (Mmm)

$$x=r \cos \theta , y=r \sin \theta$$

$$(****)=\int_{0}^{2 \pi} \int_0^{\sqrt{2}} \sqrt{ 4r^2+1}r dr d \theta= \dots= \frac{26 \pi }{6} $$

Furthermore they're using a change of variables formula for cartesian to polar, replacing $dxdy$ by $r dr d\theta$ and setting the integral boundaries as appropriate. (Wait)

 

[evinda]

Hey again! (Happy)They are using the formula:
$$\iint d \sigma= \iint \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}dA$$
where $d \sigma$ is a surface element in 3 dimensions and $dA$ is a surface element in the XY plane.
I do not believe it has a name.
If you want you can derive it, using that $\overrightarrow \nabla f$ is the (unnormalized) normal vector of the surface. (Mmm)

Furthermore they're using a change of variables formula for cartesian to polar, replacing $dxdy$ by $r dr d\theta$ and setting the integral boundaries as appropriate. (Wait)

But...isn't there also an other way to find the area of the surface..?maybe using a theorem? (Thinking)

 

[I like Serena]

But...isn't there also an other way to find the area of the surface..?maybe using a theorem? (Thinking)

Gauss's divergence theorem can sometimes be used, but I don't see how to do that here. (Sweating)

 

[evinda]

Hey again! (Happy)
They are using the formula:
$$\iint d \sigma= \iint \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}dA$$
where $d \sigma$ is a surface element in 3 dimensions and $dA$ is a surface element in the XY plane.
I do not believe it has a name.
If you want you can derive it, using that $\overrightarrow \nabla f$ is the (unnormalized) normal vector of the surface. (Mmm)

Furthermore they're using a change of variables formula for cartesian to polar, replacing $dxdy$ by $r dr d\theta$ and setting the integral boundaries as appropriate. (Wait)

So..would you use the same formula to calculate the area of the surface that is created or an other one? (Thinking) (Blush)

 

[I like Serena]

So..would you use the same formula to calculate the area of the surface that is created or an other one? (Thinking) (Blush)

Huh? (Wondering)

Which area are we talking about?

Anyway, there is another formula to calculate a surface integral.
See wiki:
$$A = \iint_S \,dS
= \iint_T \left\|{\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y}\right\| dx\, dy$$
where $\mathbf{r}(x,y)$ describes the surface $S$.

 

[evinda]

Huh? (Wondering)

Which area are we talking about?

I meant the area of the circle $x^2+y^2=2,z=2$.

 

[I like Serena]

I meant the area of the circle $x^2+y^2=2,z=2$.

Oh... yes... there is yet another formula for that one! ;)

Area of a circle disk with radius $\sqrt 2$ is $\pi (\sqrt 2)^2 = 2\pi$. (Whew)

 

[evinda]

Oh... yes... there is yet another formula for that one! ;)

Area of a circle disk with radius $\sqrt 2$ is $\pi (\sqrt 2)^2 = 2\pi$. (Whew)

But..shouldn't the result be $\displaystyle{\frac{26 \pi}{6}}$,as I have written at post #13 ? (Thinking) (Thinking)

 

[I like Serena]

But..shouldn't the result be $\displaystyle{\frac{26 \pi}{6}}$,as I have written at post #13 ? (Thinking) (Thinking)

Well... there are 2 areas that are bounded by the intersection.
A circle disk with area $2\pi$ and a part of the paraboloid with area $\dfrac{26 \pi}{6}$... (Wondering)

 

[evinda]

Well... there are 2 areas that are bounded by the intersection.
A circle disk with area $2\pi$ and a part of the paraboloid with area $\dfrac{26 \pi}{6}$... (Wondering)

I haven't really understood it.. (Sweating) Could you explain it further to me? (Thinking)

 

[I like Serena]

I haven't really understood it.. (Sweating) Could you explain it further to me? (Thinking)

I'm at a loss at what to explain. (Thinking)(Thinking)

Perhaps you can enlighten us what you do understand?
And indicate where the confusion is? (Wondering)

 

[evinda]

I'm at a loss at what to explain. (Thinking)(Thinking)

Perhaps you can enlighten us what you do understand?
And indicate where the confusion is? (Wondering)

How can we find the area of the part of the paraboloid ,which is equal to $\frac{26 \pi}{6}$ ? (Thinking)

 

[I like Serena]

How can we find the area of the part of the paraboloid ,which is equal to $\frac{26 \pi}{6}$ ? (Thinking)

For that we need to apply either the formula you gave in #13 or the formula I gave in #18. (Worried)

In effect, we need to determine what the area of an infinitesimal surface element is and integrate it.
Both formulas are derived from that concept.

See Surface integrals of scalar fields on wiki for more details how to apply #18. (Wasntme)

 

[evinda]

Huh? (Wondering)

Which area are we talking about?

Anyway, there is another formula to calculate a surface integral.
See wiki:
$$A = \iint_S \,dS
= \iint_T \left\|{\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y}\right\| dx\, dy$$
where $\mathbf{r}(x,y)$ describes the surface $S$.

What is $\mathbf{r}(x,y)$ in our case? (Thinking)

 

[I like Serena]

What is $\mathbf{r}(x,y)$ in our case? (Thinking)

That would be:
$$\mathbf{r}(x,y) = x \hat \imath + y \hat \jmath + (x^2+y^2) \hat k$$
which is a parametric expression for the paraboloid. (Smile)

 

[evinda]

That would be:
$$\mathbf{r}(x,y) = x \hat \imath + y \hat \jmath + (x^2+y^2) \hat k$$
which is a parametric expression for the paraboloid. (Smile)

I tried it now with this formula and that's what I got:

$$A=\iint_T\left \| \frac{\partial{r}}{\partial{x}} \times \frac{\partial{r}}{\partial{y}}\right \| dxdy=\iint_T \sqrt{4x^2+4y^2+1}dxdy$$

$x=r \cos{\theta}$, $y=r \sin{\theta}$

$$A=\int_0^{\sqrt{2}} \int_0^{2 \pi} \sqrt{4r^2 \cos^2{\theta}+4r^2 \sin^2{\theta}+1} r d \theta dr=\int_0^{\sqrt{2}} \int_0^{2 \pi} \sqrt{4r^2+1} r d \theta dr=\int_0^{\sqrt{2}} 2 \pi r \sqrt{4r^2+1} dr$$

$u=4r^2+1 \Rightarrow du=8dr$
$r=0 \to u=1$
$r=\sqrt{2} \to u=9$

$$A=\int_1^9 2 \pi \sqrt{u}\frac{du}{8}=\frac{\pi}{4}\int_1^9\sqrt{u}du=\frac{\pi}{4}\frac{2}{3}u^{\frac{3}{2}}|_1^9=\frac{26\pi}{6}$$

Is it right?? (Whew) (Thinking)

 

[I like Serena]

I tried it now with this formula and that's what I got:

$$A=\iint_T\left \| \frac{\partial{r}}{\partial{x}} \times \frac{\partial{r}}{\partial{y}}\right \| dxdy=\iint_T \sqrt{4x^2+4y^2+1}dxdy$$

$x=r \cos{\theta}$, $y=r \sin{\theta}$

$$A=\int_0^{\sqrt{2}} \int_0^{2 \pi} \sqrt{4r^2 \cos^2{\theta}+4r^2 \sin^2{\theta}+1} r d \theta dr=\int_0^{\sqrt{2}} \int_0^{2 \pi} \sqrt{4r^2+1} r d \theta dr=\int_0^{\sqrt{2}} 2 \pi r \sqrt{4r^2+1} dr$$

$u=4r^2+1 \Rightarrow du=8dr$
$r=0 \to u=1$
$r=\sqrt{2} \to u=9$

$$A=\int_1^9 2 \pi \sqrt{u}\frac{du}{8}=\frac{\pi}{4}\int_1^9\sqrt{u}du=\frac{\pi}{4}\frac{2}{3}u^{\frac{3}{2}}|_1^9=\frac{26\pi}{6}$$

Is it right?? (Whew) (Thinking)

Yep. All correct. (Smile)

Erm... except for the $du=8dr$, which should be $du=8rdr$.
Apparently that is what you used anyway. (Mmm)

 

[evinda]

Yep. All correct. (Smile)

Erm... except for the $du=8dr$, which should be $du=8rdr$.
Apparently that is what you used anyway. (Mmm)

Yes,that's what I used.. (Nod)Thank you very much! (Mmm) (Smirk)