# What are the limits for integrating a constrained surface with two variables?

In summary, you can calculate the normal vector by: dr/du \times dr/dv and use limits to integrate only within the elipse.
Homework Statement
Calculate the surface of the plane
$$ax + by + c = d$$
inside the elipse
$$x^2/a^2 + y^2/b^2 = 1$$
Relevant Equations
Surface integrals
I start by parametarize the surface with two variables:
$$r(u,v) = (u, v, \frac {d -au -bv} c)$$

The I can get the normal vector by
$$dr/du \times dr/dv$$

What limits should I use to integrate this only within the elipse?

I could redo the whole thing and try write r(u, v) as u being the radius percentage and v being the angle such that
$$r(u, v) = (uacos(v), ubsin(v), (d - au^2cos(v) - ub^2sin(v))/c)$$
$$u: 0 \rightarrow 1, v: 0 \rightarrow 2\pi$$
But good luck calculating the cross product, even worse: the absolute value of the cross product.

It's a billion numbers, I am certain that is not the correct way to solve it.

Last edited:
Homework Statement:: Calculate the surface of the plane
$$ax + by + c = d$$

I assume $ax + by + cz = d$ is meant.

inside the elipse
$$x^2/a^2 + y^2/b^2 = 1$$
Relevant Equations:: Surface integrals

I start by parametarize the surface with two variables:
$$r(u,v) = (u, v, \frac {d -au -bv} c)$$

The I can get the normal vector by
$$dr/du \times dr/dv$$

What limits should I use to integrate this only within the elipse?

I could redo the whole thing and try write r(u, v) as u being the radius percentage and v being the angle such that
$$r(u, v) = (racos(v), rbsin(v), (d - ar^2cos(v) - rb^2sin(v))/c)$$
$$u: 0 \rightarrow 1, v: 0 \rightarrow 2\pi$$
But good luck calculating the cross product, even worse: the absolute value of the cross product.

It's a billion numbers, I am certain that is not the correct way to solve it.

Your notation is a bit confusing here; you seem confused as to whether your radial parameter is $r$ or $u$, and I don't think you calculated $z$ correctly.

Keep going. This is unquestionably the correct approach. The standard parametrization of the interior of this ellipse is $(x,y) = (ar\cos\theta,br\sin\theta)$ with $0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$. The cross-product is straightforward, if tedious, to calculate and many terms will either cancel or simplify using $\cos^2\theta + \sin^2\theta = 1$.

The first term of the normal vector is
$$n_x = abu^2/c*sin(v)^2 -ub^3/c*cos(v)*sin(v) - 2abu^2/c*cos(v)^2 + r^2b^2*sin(v)cos(v)$$
and I'm suppose to square that along with 2 other terms??

I rather just quit school all together.

The first term of the normal vector is
$$n_x = abu^2/c*sin(v)^2 -ub^3/c*cos(v)*sin(v) - 2abu^2/c*cos(v)^2 + r^2b^2*sin(v)cos(v)$$
and I'm suppose to square that along with 2 other terms??

I rather just quit school all together.

You haven't calculated $z = (d - a^2u\cos v - b^2 u\sin v)/c$ correctly. I find $$\begin{split} n_x &= -\frac{ub\sin v}{c}\left(-a^2 \sin v + b^2\cos v\right) + \frac{ub\cos v}{c}\left( a^2\cos v +b^2 \sin v\right) \\ &= \frac{ba^2 u}{c}.\end{split}$$ However, it has occurred to me that the easiest way is to calculate $\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}$ using $(x,y,z) = (u, v, (d - au - bv)/c)$ and only then make the change of variable $(u,v) = (ar\cos\theta, br\sin\theta)$.

I tried your approach, but even simpler I calculated ##r_u \times r_v ## then I tried to do the replacement.
$$k = r_u \times r_v = \sqrt{A^2 + B^2 + C^2}/C^2$$
To get the surface area I do:
$$\iint |r_u \times r_v| du dv = k \iint du dv$$

Now the issue is convering the final integral to ##d\phi dr##
I can't simply find du by derivation since ##u = racos(\phi)##, I'd get two partial derivatives.

So I'm not sure how to procced from here.

I redid it and as you say, all the cosine and sine cancels out and you end up with just abc combinations, making calculating the absolute value much easier.

Redid and got the correct answer!
But this sure was an exercise in pedantry and nothing else.

## 1. What is a constrained surface integral?

A constrained surface integral is a mathematical concept used in multivariable calculus to calculate the flux of a vector field across a surface. It takes into account the orientation and curvature of the surface, as well as any constraints or boundaries that may be present.

## 2. How is a constrained surface integral different from a regular surface integral?

A regular surface integral calculates the flux of a vector field across a surface without taking into account any constraints or boundaries. A constrained surface integral, on the other hand, considers these factors and provides a more accurate calculation of the flux.

## 3. What are some real-world applications of constrained surface integrals?

Constrained surface integrals are commonly used in physics and engineering to calculate the flow of a fluid across a surface, such as the flow of air over an airplane wing. They are also used in electromagnetism to calculate the electric or magnetic flux across a surface.

## 4. How is a constrained surface integral calculated?

To calculate a constrained surface integral, the surface must first be parameterized and the vector field must be defined. Then, the integral is evaluated using a double integral over the parameterized surface, taking into account any constraints or boundaries.

## 5. Are there any limitations or challenges when using constrained surface integrals?

One limitation of constrained surface integrals is that they can be difficult to visualize and calculate for complex surfaces. Additionally, finding the appropriate surface parameterization and setting up the integral can be challenging. However, with practice and familiarity, these challenges can be overcome.

• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
878
• Calculus and Beyond Homework Help
Replies
3
Views
689
• Calculus and Beyond Homework Help
Replies
24
Views
3K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
997
• Calculus and Beyond Homework Help
Replies
1
Views
647
• Calculus and Beyond Homework Help
Replies
21
Views
1K
• Calculus and Beyond Homework Help
Replies
21
Views
2K