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

[Addez123]

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.

 

[pasmith]

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 $\cos^2\theta + \sin^2\theta = 1$.

 

[Addez123]

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.

 

[pasmith]

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 $(u,v) = (ar\cos\theta, br\sin\theta)$.

 

[Addez123]

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.

 

[Addez123]

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.