Calculate the volume of the space D

[mathmari]

Hey!

I want to draw the space $D$ that is between the paraboloid with equation $z=x^2+y^2$ and the cone with equation $z=\sqrt{x^2+y^2}$ and I want to calculate its volume. How does $D$ look like?

To calculate the volume we have to calculate the integral $\iiint_D dxdydz$.

I have done the following:

Let $x^2+y^2=t^2$, at the intersection of the two surfaces we have $t^2=t\Rightarrow t^2-t=0\Rightarrow t=0 , t=1$.

For $t=0$ we have $x^2+y^2=0\Rightarrow x=y=0$.

For $t=1$ we have $x^2+y^2=1\Rightarrow y^2=1-x^2\Rightarrow -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}$. With $y=0$ we get $0\leq y \leq \sqrt{1-x^2}$. So that the square root is defined it must hold $1-x^2\geq 0\Rightarrow x^2\leq 1\Rightarrow -1\leq x\leq 1$. With $x=0$ we get $0\leq x\leq 1$.

Is everything correct so far? But which are the bounds for $z$? Is it maybe $\sqrt{x^2+y^2}\leq z\leq x^2+y^2$? (Wondering)

 

[I like Serena]

Hey!

I want to draw the space $D$ that is between the paraboloid with equation $z=x^2+y^2$ and the cone with equation $z=\sqrt{x^2+y^2}$ and I want to calculate its volume.

How does $D$ look like?

Hey mathmari! (Smile)

To draw a surface or volume, the usual way to do it, is to calculate the curves that intersect the xy-plane, the xz-plane, and the yz-plane, and draw those.
In our case we have 2 surfaces that intersect each other, so we should find their intersection curves and draw those as well.
Can we find and draw those? (Wondering)

For $t=1$ we have $x^2+y^2=1\Rightarrow y^2=1-x^2\Rightarrow -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}$. With $y=0$ we get $0\leq y \leq \sqrt{1-x^2}$. So that the square root is defined it must hold $1-x^2\geq 0\Rightarrow x^2\leq 1\Rightarrow -1\leq x\leq 1$. With $x=0$ we get $0\leq x\leq 1$.

I'm not sure what you mean with 'With $y=0$ we get $0\leq y \leq \sqrt{1-x^2}$'?

Anyway, I'd recommend to look at it from cylindrical coordinates.
At $t=1$, or rather $\rho=\sqrt t=1$ (radius parallel to the xy-plane), the paraboloid is a circle with radius 1, and the cone is also a circle with radius 1.

Is everything correct so far? But which are the bounds for $z$? Is it maybe $\sqrt{x^2+y^2}\leq z\leq x^2+y^2$? (Wondering)

How would it look in cylindrical coordinates, say, $(\rho,\phi,z)$? (Wondering)

 

[mathmari]

I'm not sure what you mean with 'With $y=0$ we get $0\leq y \leq \sqrt{1-x^2}$'?

I thought so since for $t=0$ we get $y=0$, and so the space is bounded by $y=0$. Is this wrong?

Anyway, I'd recommend to look at it from cylindrical coordinates.

Can we not use cartesian coordinates? (Wondering)

 

[I like Serena]

I thought so since for $t=0$ we get $y=0$, and so the space is bounded by $y=0$. Is this wrong?

We found that the intersection of the paraboloid and the cone consists of the point $(0,0,0)$, and separate from that, the curve given by $x^2+y^2=1, z=1$.
Somehow they must bound our volume.
It helps to draw them to figure out how.
From my own drawing I'm concluding that the ranges for $x$ and $y$ that you found are not correct. (Shake)

Can we not use cartesian coordinates?

Sure.
I was effectively only suggesting to define $\rho=\sqrt{x^2+y^2}$ as an intermediate step.
Anyway, I believe we should get $D=\{(x,y,z) \mid 0\le z \le 1,\ z^2 \le x^2+y^2 \le z\}$. (Thinking)

But which are the bounds for $z$? Is it maybe $\sqrt{x^2+y^2}\leq z\leq x^2+y^2$?

I believe it should be the other way around:
$$x^2+y^2 \le z \le \sqrt{x^2+y^2}$$
(Thinking)

 

[HallsofIvy]

In the y= 0 plane, the paraboloid $z= x^2+ y^2$ becomes the parabola $z= x^2$ and the cone $z= \sqrt{x^2+ y^2}$ becomes $z= \sqrt{x^2}= |x|$, a pair of lines. Which is higher?

 

[mathmari]

Anyway, I believe we should get $D=\{(x,y,z) \mid 0\le z \le 1,\ z^2 \le x^2+y^2 \le z\}$. (Thinking)
I believe it should be the other way around:
$$x^2+y^2 \le z \le \sqrt{x^2+y^2}$$
(Thinking)

Ah ok.

How could we bound the variables $x$ and $y$ ? Do we have the following? (Wondering)

We have that $z^2 \le x^2+y^2 \le z$, since $z\leq 1$, we get that $x^2+y^2\leq 1\Rightarrow x^2\leq 1-y^2 \Rightarrow -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}$.
So that the square root is defined it must hold $1-y^2\geq 0 \Rightarrow y^2\leq 1 \Rightarrow -1\leq y \leq 1$.

Are these bounds correct? (Wondering)

In the y= 0 plane, the paraboloid $z= x^2+ y^2$ becomes the parabola $z= x^2$ and the cone $z= \sqrt{x^2+ y^2}$ becomes $z= \sqrt{x^2}= |x|$, a pair of lines. Which is higher?

For $x\in [0,1]$ we have that $|x|$ is higher than $ x^2$. That means that the same holds also when we don't consider the plane $y=0$, right? (Wondering)

 

[I like Serena]

Ah ok.

How could we bound the variables $x$ and $y$ ? Do we have the following? (Wondering)

We have that $z^2 \le x^2+y^2 \le z$, since $z\leq 1$, we get that $x^2+y^2\leq 1\Rightarrow x^2\leq 1-y^2 \Rightarrow -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}$.
So that the square root is defined it must hold $1-y^2\geq 0 \Rightarrow y^2\leq 1 \Rightarrow -1\leq y \leq 1$.

Are these bounds correct?

Yep.

 

[mathmari]

Yep.

Great! Thank you! (Mmm)

 

[Prove It]

Ah ok.

How could we bound the variables $x$ and $y$ ? Do we have the following? (Wondering)

We have that $z^2 \le x^2+y^2 \le z$, since $z\leq 1$, we get that $x^2+y^2\leq 1\Rightarrow x^2\leq 1-y^2 \Rightarrow -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}$.
So that the square root is defined it must hold $1-y^2\geq 0 \Rightarrow y^2\leq 1 \Rightarrow -1\leq y \leq 1$.

Are these bounds correct? (Wondering)


For $x\in [0,1]$ we have that $|x|$ is higher than $ x^2$. That means that the same holds also when we don't consider the plane $y=0$, right? (Wondering)

Of course, since every cross-section parallel to the x-y plane will be circles (or rather, toruses), the integral will be infinity times easier to solve with cylindrical polar co-ordinates.

You have already found $\displaystyle \begin{align*} z^2 \leq x^2 + y^2 \leq z \implies z^2 \leq r \leq z \end{align*}$, with $\displaystyle \begin{align*} 0 \leq z \leq 1 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\,\pi \end{align*}$, thus the volume is

$\displaystyle \begin{align*} V &= \int_0^{2\,\pi}{ \int_0^1{\int_{z^2}^z{ r\,\mathrm{d}r }\,\mathrm{d}z} \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \int_0^1{ \left[ \frac{r^2}{2} \right] _{z^2}^z \,\mathrm{d}z } \,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\,\pi}{ \int_0^1{ \left( z^2 - z^4 \right) \,\mathrm{d}z } \,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\,\pi}{ \left[ \frac{z^3}{3} - \frac{z^5}{5} \right] _0^1 \,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\,\pi}{ \left( \frac{1}{3} - \frac{1}{5} \right) \,\mathrm{d}\theta } \\ &= \frac{1}{2} \cdot \frac{2}{15} \int_0^{2\,\pi}{ 1\,\mathrm{d}\theta } \\ &= \frac{1}{15} \, \left[ \theta \right] _0^{2\,\pi} \\ &= \frac{2\,\pi}{15}\,\textrm{units}^3 \end{align*}$

 

[mathmari]

Of course, since every cross-section parallel to the x-y plane will be circles (or rather, toruses), the integral will be infinity times easier to solve with cylindrical polar co-ordinates.

You have already found $\displaystyle \begin{align*} z^2 \leq x^2 + y^2 \leq z \implies z^2 \leq r \leq z^2 \end{align*}$, with $\displaystyle \begin{align*} 0 \leq z \leq 1 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\,\pi \end{align*}$, thus the volume is

$\displaystyle \begin{align*} V &= \int_0^{2\,\pi}{ \int_0^1{\int_{z^2}^z{ r\,\mathrm{d}r }\,\mathrm{d}z} \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \int_0^1{ \left[ \frac{r^2}{2} \right] _{z^2}^z \,\mathrm{d}z } \,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\,\pi}{ \int_0^1{ \left( z^2 - z^4 \right) \,\mathrm{d}z } \,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\,\pi}{ \left[ \frac{z^3}{3} - \frac{z^5}{5} \right] _0^1 \,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\,\pi}{ \left( \frac{1}{3} - \frac{1}{5} \right) \,\mathrm{d}\theta } \\ &= \frac{1}{2} \cdot \frac{2}{15} \int_0^{2\,\pi}{ 1\,\mathrm{d}\theta } \\ &= \frac{1}{15} \, \left[ \theta \right] _0^{2\,\pi} \\ &= \frac{2\,\pi}{15}\,\textrm{units}^3 \end{align*}$

I see! Thank you very much! (Smile)