Surface area of a plane inside a cone

Question

Determine the surface area of the part of the plane $z=1+x+2y$ which is inside the cone surface $z=\sqrt{2x^{2}+10y^{2}}$.

Answer 1

Better check me on this, but I think this is the way to go.

First of all, luckily, you're integrating on a plane, even if it's a tilted plane. So the question is, given a $dx$ and a $dy$, what is the corresponding $dA$ on that plane. Since $z=1+x+2y$, the vector corresponding to $x$ is really $\hat{x}+\hat{z}$ and the vector corresponding to $y$ is $2\hat{y}+\hat{z}$. You can get $dA$ by taking the cross_product, which turns out to be $d\vec{A}=(\hat{z}-2\hat{y}-\hat{x})dxdy$. Take the magnitude and $dA=\sqrt{6}dxdy$. This is good news. It means that you can project your tilted plane onto the $x-y$ plane, do your area integral there, and then multiply by $\sqrt{6}$ to get the area in the actual tilted plane.

So what do you integrate over in the $x-y$ plane?

We know $1+x+2y = z = \sqrt{2x^2 + 10y^2}$, so eliminate $z$ to get

$(1+x+2y)^2 = 2x^2 + 10y^2$

Now, this is the equation of a tilted ellipse in the $x-y$ plane. The trick is to get it in standard ellipse form so you can just read the area right out of the formula. Unfortunately, I'm going to have to leave you here. I don't remember and can't find the formula for doing this in the general case. I suspect that this is actually an easy case -- a tilt of 30 degrees or 45 degrees or what have you -- but I'm not seeing it.

Answer 2

Wow...Well this is definitely one of the toughest questions II have come across...and it piqued my curiosity.

So, to begin with, you have two equations.

$1:z=1+x+2y$ $2:z=\sqrt{2x^2+10y^2}$

You would want to find the area which the two intersect in, first...that would be the area which both are equal to each other.

That would be when: $1+x+2y=\sqrt{2x^2+10y^2}$

After a bit of simplifying, you'll realize that this is the ellipse defined by the implicit equation: $6y^2-4y-4yx-2x-1+x^2=0$ $(6)y^2+(-4-4x)y+(x^2-2x-1)=0$

Using the quadratic formula making your a,b, and c values the x components, yields the equation of the ellipse in two components: $$\dfrac{2+2x\pm\sqrt{10x+5-x^2}}{6}$$

This is your area of surface integration.

Now as for the integrand. When you take the function: $$f(x,y)=1+x+2y$$ This can be parameterized to $$f(x,y)=(x,y,1+x+2y)$$

Let $r=\dfrac{\partial f}{\partial x}$ Let $s=\dfrac{\partial f}{\partial y}$

The integrand for surface area is defined as $$\Arrowvert r \times s \Arrowvert $$

So:

$r=(1,0,1)$

$s=(0,1,2)$

Let $p=(r \times s)$

$p=(r \times s)=(-1,-2,1)$

$$\Arrowvert p \Arrowvert=\sqrt{(-1)^2+(-2)^2+(1)^2}$$ $$=\sqrt{1+4+1}$$ $$=\sqrt{6}$$

Therefore our integrand is $\sqrt{6}$

One can define the limits of integration for the integrand by stating the x-interval as the domain of the ellipse and the y-interval as the upper and lower ellipse function:

Let:

$m=5-\sqrt{30}$

$n=5+\sqrt{30}$

$f(x)=\dfrac{2+2x-\sqrt{10x+5-x^2}}{6}$

$g(x)=\dfrac{2+2x+\sqrt{10x+5-x^2}}{6}$

Therefore the integration limits are

$m<x<n$

$f(x)<y<g(x)$

Let $R=[m,n] \centerdot [f(x),g(x)]$

Let $\sigma$ represent surface area

So:

$$\sigma = \iint_R{\sqrt{6}} \ dA $$ $$=\int_m^n{ [ \int_{f(x)}^{g(x)}{\sqrt{6}} \ dy ] } \ dx$$ $$=\int_m^n{\gamma} \ dx$$

$\gamma =\int_{f(x)}^{g(x)}{\sqrt{6}} \ dy$

$=\left(\sqrt{6} y \right) \arrowvert_{f(x)}^{g(x)}$

$=\sqrt{6}[(f-g)(x)]$

$=\sqrt{6}(\dfrac{2+2x+\sqrt{10x+5-x^2}}{6} - \dfrac{2+2x-\sqrt{10x+5-x^2}}{6})$

$=\sqrt{6}(2 \dfrac{\sqrt{10x+5-x^2}}{6})$

$=\dfrac{2 \sqrt{6}}{6} \sqrt{10x+5-x^2}$

$$ \sigma =\int_m^n{\dfrac{2 \sqrt{6}}{6} \sqrt{10x+5-x^2}}$$

$$= \int_{5- \sqrt{30}}^{5+ \sqrt{30}} { \dfrac{2 \sqrt{6}}{6} \sqrt{10x+5-x^2}} $$

$$= \dfrac{2 \sqrt{6}}{6} \ \int_{5-\sqrt{30}}^{5+\sqrt{30}}{\sqrt{10x+5-x^2}} $$

$$\sigma =\dfrac{2 \sqrt{6}}{6} \lambda$$

$$ \lambda = \int_{5-\sqrt{30}}^{5+\sqrt{30}}{\sqrt{10x+5-x^2}}$$

(... http://m.wolframalpha.com/input/?i=integrate+sqrt%2810x%2B5-x%5E2%29+from+-0.477%3Cx%3C10.477&x=0&y=0)

$$ \lambda \approx 47.1239$$

$$\sigma \approx \dfrac{2 \sqrt{6}}{6} (47.1239)$$

$$\sigma \approx 38.4765$$

$\therefore$ The surface area is about $38.4765 units^2$

Well, there it is!

If I were you though, I would check this answer before hand just to make sure it is right

Or respond if it was the correct answer please so I know!

Answer 3

The surface we are interested in is an ellipse $E$. We project $E$ to the $(x,y)$-plane by eliminating $z$ from the two equations $$z^2=2x^2 +10 y^2, \quad z=1+x+2y\ .$$ One gets $$x^2-4xy+6y^2-2x-4y-1=0\ .$$ We now write $x=x_0+u$, $\>y=y_0+v$ and determine $(x_0,y_0)$ (the center of the ellipse) such that the resulting equation in $u$ and $v$ has no linear terms. One finds $(x_0,y_0)=(5,2)$, and the resulting equation is $$u^2-4uv+6v^2=10\ .$$ Using Lagrange's method one now can determine the semiaxes $a$ and $b$ of this projected ellipse $E'$, and then $A(E')=\pi a b$.

Now the unit normal of the cutting plane is $n={1\over\sqrt{6}}(-1,-2,1)$; so $n$ is tilted with respect to the $z$-axis by an angle $\theta$ with $\cos\theta={1\over\sqrt{6}}$. It follows that the area of the original ellipse $E$ is given by $$A(E)={1\over\cos\theta}A(E')=\pi\sqrt{6}\>ab\ .$$ I leave the computation of $a$ and $b$ to you.