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

In summary, we have a conversation discussing how to calculate the area of a conic surface defined by the relations $z^2=x^2+y^2$ and $a\leq z\leq b$. The first method discussed is using cartesian coordinates, where we determine the boundaries of the region and use the function $\Sigma(x,y)=(x,y,\sqrt{x^2+y^2})$ to find the area. The second method discussed is using cylindrical coordinates, where we use the function $\Sigma(r,\theta)=(r,\theta,r)$ and integrate over the region $D=[a,b]\times [0,2\pi]$. Both methods require further calculations and integration to find the final area.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

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)
 
Physics news on Phys.org
  • #2
mathmari said:
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)

mathmari said:
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)

mathmari said:
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)
 
  • #3
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)
 

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

What is the formula for finding the area of a conic surface?

The formula for finding the area of a conic surface depends on the type of conic. For a circle, the formula is A=πr^2, where r is the radius. For an ellipse, the formula is A=πab, where a and b are the semi-major and semi-minor axes, respectively. For a parabola, the formula is A=2/3bh, where b is the base and h is the height. And for a hyperbola, the formula is A=πab, where a and b are the semi-major and semi-minor axes, respectively.

How does the shape of a conic surface affect its area?

The shape of a conic surface directly affects its area. The formula for finding the area takes into account the specific shape of the conic, so a change in shape will result in a different area. For example, a circle and an ellipse may have the same radius, but their areas will be different because of their different shapes.

Can the area of a conic surface be negative?

No, the area of a conic surface cannot be negative. The formula for finding the area only yields positive values, and the area of a shape cannot be negative. However, the area of a conic surface may be imaginary for certain values of the variables in the formula, such as for a hyperbola with a negative value for the semi-minor axis.

Is the area of a conic surface affected by its orientation?

Yes, the orientation of a conic surface can affect its area. For example, a circle with a horizontal orientation will have a different area than the same circle with a vertical orientation. This is because the orientation can change the values of the variables in the formula for finding the area.

How can the area of a conic surface be used in real life?

The area of a conic surface has many real-life applications. For example, it is used in engineering and architecture for designing structures such as bridges and tunnels. It is also used in physics and astronomy for calculating the surface area of planets and other celestial bodies. In everyday life, the area of a conic surface is used in cooking and baking, where the area of a circle is needed for calculating the size of a pizza or cake.

Similar threads

Replies
9
Views
2K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
10
Views
3K
Replies
5
Views
2K
Replies
10
Views
2K
Back
Top