MHB How can the area of a conic surface be calculated in two different ways?

[mathmari]

Hey!

If $0\leq a<b$, I want to calculate the area of the conic surface that is defined by the relations $z^2=x^2+y^2$ and $a\leq z\leq b$ in two ways:

1. using cartesian coordinates
2. using cylindrical coordinates

I have the following:

1. We have that $z^2=x^2+y^2\Rightarrow z=\pm \sqrt{x^2+y^2}$. Since $z\geq a\geq 0$ we have that $z=\sqrt{x^2+y^2}$.
   We have that $$a\leq z\leq b\Rightarrow a\leq \sqrt{x^2+y^2}\leq b \Rightarrow a^2\leq x^2+y^2\leq b^2 \Rightarrow a^2-x^2\leq y^2\leq b^2-x^2 \Rightarrow \sqrt{a^2-x^2}\leq y\leq \sqrt{b^2-x^2}$$
   I am not really sure if we can just take the root of the inequality. (Wondering)
   
   If this is correct, then the square roots are defined if $a^2-x^2\geq 0$ and $b^2-x^2\geq 0$. Since $a<b$ we get that $-a\leq x\leq a$.
   
   So, we define the function $\Sigma : D \rightarrow \mathbb{R}^3$ with $\Sigma (x,y)=(x,y, x^2+y^2)$, where $D=[-a,a]\times[\sqrt{a^2-x^2}, \sqrt{b^2-x^2}]$.
   
   We have that $\Sigma_x=(1,0,2x)$ and $\Sigma_y=(0,1,2y)$.
   
   So, we get $\Sigma_x\times\Sigma_y=(-2x, -2y, 1)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{4x^2+4y^2+1}$.
   
   So, we get $$A(\Sigma (D))=\iint_D\|\Sigma_x\times\Sigma_y\|dxdy=\int_{-a}^a\int_{\sqrt{a^2-x^2}}^{\sqrt{b^2-x^2}}\sqrt{4x^2+4y^2+1}dydx$$ Is everything correct so far? If yes, how could we continue? (Wondering)

 

[I like Serena]

If $0\leq a<b$, I want to calculate the area of the conic surface that is defined by the relations $z^2=x^2+y^2$ and $a\leq z\leq b$

We have that $z^2=x^2+y^2\Rightarrow z=\pm \sqrt{x^2+y^2}$. Since $z\geq a\geq 0$ we have that $z=\sqrt{x^2+y^2}$.
We have that $$a\leq z\leq b\Rightarrow a\leq \sqrt{x^2+y^2}\leq b \Rightarrow a^2\leq x^2+y^2\leq b^2 \Rightarrow a^2-x^2\leq y^2\leq b^2-x^2 \Rightarrow \sqrt{a^2-x^2}\leq y\leq \sqrt{b^2-x^2}$$
I am not really sure if we can just take the root of the inequality. (Wondering)

Hey mathmari! (Smile)

Not just like that. We have to consider cases.
We can already find that $-b\le x \le b$ (can we?), so $b^2-x^2 \ge 0$.
However, we can either have $a^2-x^2<0$ or $a^2-x^2\ge 0$

In the first case we always have $a^2-x^2 \le y^2$, since $y^2 \ge 0$.
So we can leave out the left side - it's always satisfied.
Then we can take the square root, but we get $|y| \le \sqrt{b^2-x^2}$ with absolute signs.

In the second case we get $\sqrt{a^2-x^2}\leq |y|\leq \sqrt{b^2-x^2}$, again with absolute signs.

It needs a little more work to find the boundaries. (Thinking)

So, we define the function $\Sigma : D \rightarrow \mathbb{R}^3$ with $\Sigma (x,y)=(x,y, x^2+y^2)$, where $D=[-a,a]\times[\sqrt{a^2-x^2}, \sqrt{b^2-x^2}]$.

I'm afraid a cartesian product of 2 coordinates always yields a rectangular region.
So the second coordinate cannot depend on the first coordinate $x$.
We'll need a different notation to define $D$. (Worried)

We have that $\Sigma_x=(1,0,2x)$ and $\Sigma_y=(0,1,2y)$.

So, we get $\Sigma_x\times\Sigma_y=(-2x, -2y, 1)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{4x^2+4y^2+1}$.

So, we get $$A(\Sigma (D))=\iint_D\|\Sigma_x\times\Sigma_y\|dxdy=\int_{-a}^a\int_{\sqrt{a^2-x^2}}^{\sqrt{b^2-x^2}}\sqrt{4x^2+4y^2+1}dydx$$ Is everything correct so far? If yes, how could we continue? (Wondering)

Leaving out the boundaries for now, since they need some work, we have:
$$A=\iint_D \sqrt{4x^2+4y^2+1}dydx$$
To find the innermost integral, we can use the following standard integral:
$$\int \sqrt{c^2 + u^2}du = \frac 12(u\sqrt{c^2+u^2} + c^2\ln(u+\sqrt{c^2+u^2})) + C$$
That is for the first integration... and then we have to integrate again! (Sweating)

 

[I like Serena]

Wait! I found a rather important mistake. (Wait)

We should have $\Sigma(x,y)=(x,y,\sqrt{x^2+y^2})$.
Consequently we get:
$$\|\Sigma_x\times \Sigma_y\| = \sqrt 2$$

That is a bit easier to integrate! (Whew)

Btw, I believe the domain $D$ in cartesian coordinates should be:
$$
D=\{ (x,y)\in \mathbb R^2: a<|x|\le b \land |y|\le \sqrt{b^2-x^2} \}
\cup \{(x,y)\in \mathbb R^2: |x|\le a \land \sqrt{a^2-x^2} \le |y| \le \sqrt{b^2-x^2}\}
$$

Then the cartesian integral becomes:
$$
A(\Sigma(D)) \\
= \int_{-b}^{-a} \int_{-\sqrt{b^2-x^2}}^{\sqrt{b^2-x^2}} \sqrt 2\,dydx
+ \int_{-a}^{a} \int_{-\sqrt{b^2-x^2}}^{-\sqrt{a^2-x^2}} \sqrt 2\,dydx
+ \int_{-a}^{a} \int_{\sqrt{a^2-x^2}}^{\sqrt{b^2-x^2}} \sqrt 2\,dydx
+ \int_{a}^{b} \int_{-\sqrt{b^2-x^2}}^{\sqrt{b^2-x^2}} \sqrt 2\,dydx \\
= 2\int_{a}^{b} \int_{-\sqrt{b^2-x^2}}^{\sqrt{b^2-x^2}} \sqrt 2\,dydx
+ 2\int_{-a}^{a} \int_{\sqrt{a^2-x^2}}^{\sqrt{b^2-x^2}} \sqrt 2\,dydx
$$
(Thinking)

In cylindrical coordinates this becomes:
$$\Sigma: D\to \mathbb R^3\text{ given by }\Sigma(r,\theta)=(r,\theta,r)$$
where:
$$D=[a,b]\times [0,2\pi]$$
and:
$$A(\Sigma(D))=\int_a^b \int_0^{2\pi} \sqrt 2\ r\,d\theta\,dr$$
(Thinking)