# Area of the intersection of a cone and a sphere

Calculate the area of the surface $x^2+y^2 = z^2$ with $z \geq 0$, limited by $x^2 +y^2+z^2=2ax$. I think the method to solve it is to calculate the parametric equation of the curve and then calculate the area of the enclosed curve using $\frac{1}{2} \int_c x\,dy - y\,dx$. I'm having trouble finding the equation of the parametric curve. Thanks.

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Welcome to MSE! What have tried? Where are you stuck? Is this HW? If so, it should be tagged as such. Regards –  Amzoti Jan 22 '13 at 23:21
A parametric equation for the intersection curve won't help, I think, since a given closed curve in space may be the bounding curve of a large family of surfaces having various areas bounded by that curve. As I recall the formula $(1/2)\int_C(xdy-ydx)$ is for curves lying in the $xy$ plane. –  coffeemath Jan 23 '13 at 11:36

Easy way:

The projection of the "intersection" on the x-y plane is a circle of radius $\frac{a}{2}$. Since the normals on the cone is making an $45^\circ$ angle with the unit z-vector, the area of the intersection is $\frac{\pi}{4} a^2 / \cos{45^\circ} = \frac{\pi}{2 \sqrt{2}} a^2$.

Harder? way:

Introduce parametrization $(r,\theta) \rightarrow (\frac{r}{\sqrt{2}} \cos{\theta}, \frac{r}{\sqrt{2}} \sin{\theta}, \frac{r}{\sqrt{2}} )$ to the cone surface.
The cone and sphere intersect at: $$r^2 = x^2 + y^2 + z^2 = 2ax = \sqrt{2} a r \cos{\theta} \implies -\pi/2 \le \theta \le \pi/2 \text{ and } 0 \le r \le \sqrt{2} a \cos{\theta}.$$

In terms of $(r,\theta)$, the metric is $dx^2 + dy^2 + dz^2 = dr^2 + \frac{r^2}{2} d\theta^2$, the surface area element is $\sqrt{\frac{r^2}{2}} dr d\theta$ and the area we want becomes:

$$\frac{1}{\sqrt{2}} \int_{-\pi/2}^{\pi/2} d\theta \int_0^{\sqrt{2} a \cos{\theta}} r dr = \frac{a^2}{\sqrt{2}} \int_{-\pi/2}^{\pi/2} \cos{\theta}^2 d\theta = \frac{\pi}{2 \sqrt{2}} a^2$$

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Another way to look at it: as coffemath says, your surface is the part of the cone $x^2+y^2 = z^2$ inside the sphere $(x-a)^2 + y^2 + z^2 = a^2$ (with $z>0$). So you can parametrize your surface as

$$\varphi (x,y) = \left( x, y, \sqrt{x^2+y^2} \right) \qquad \text{with} \qquad (x,y) \in D$$

where $D$ is the projection of your part of the cone onto the $xy$ plane, which is the circle

$$D = \left\{(x,y) \ \vert \ \left(x-\frac{a}{2} \right)^2 + y^2 \leq \left( \frac{a}{2} \right)^2 \right\} \ .$$

Now to put this as limits of your integrals doesn't look too much nice to me. Instead, if you resort to polar coordinates,

$$x = r\cos\theta , \quad y = r\sin\theta \ ,$$

you can describe this same cercle as

$$D = \left\{(r,\theta)\ \vert \ \theta \in [0,\pi] \ , \quad 0 \leq r \leq a\sin\theta \right\} \ ,$$

which looks easier to me as limits of integration.

Exercise 1. Check this description of $D$ with polar coordinates.

Of course, you have to change your parametrization to polar coordinates too:

$$\varphi (r,\theta ) = (r\cos\theta , r\sin\theta , r)$$

and now compute:

$$\int_D ||\varphi_r \times \varphi_\theta|| drd\theta = 2\int_0^{\pi/2}\int_0^{a\sin\theta}r\sqrt{2}drd\theta \ .$$

Exercise 2. Finish the computation of this integral. :-)

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The cone has equation [1] $x^2+y^2=z^2$ while the sphere equation is [2] $x^2+y^2+z^2=2ax.$ I'll assume $a>0$ since by symmetry this won't change the area compared to the case $a<0$.

Using [1], equation [2] becomes $2z^2=2ax$, i.e. $z=\sqrt{a}\cdot \sqrt{x}.$ Now using $z^2=ax$ in equation [1] gives

[3] $x^2+y^2=ax$,

this being the projection of the part of the cone you seek the area of onto the $xy$ plane. Equation [3] is that of a circle in the $xy$ plane, center $(a/2,0)$ radius $a/2$. Now you want to use that the area of a surface $z=f(x,y)$ over a region $R$ in the $xy$ plane is obtained by integrating $\sqrt{1+z_x^2+z_y^2}$ over $R$, where $z_x,\ z_y$ are the partial derivatives.

In your example, since $z=\sqrt{a}\cdot \sqrt{x},$ the $y$ partial is zero and you're integrating $\sqrt{1+a/(4x)}$ over the region bounded by the circle described by equation [3]. So set this integral up as an iterated integral, with $x$ (the outer iteration) going from $0$ to $a$ and $y$ (the inner iteration) going from $-\sqrt{(a/2)^2-(x-a/2)^2}$ to $+\sqrt{(a/2)^2-(x-a/2)^2}.$

EDIT: I used the wrong formula for what to integrate above. Instead of $z=\sqrt{a}\cdot \sqrt{x},$ one should use $z=\sqrt{x^2+y^2}$ which comes right from the cone formula. The other formula is only valid while finding the intersection of the cone and sphere. So the thing to integrate is $\sqrt{1+z_x^2+z_y^2}$, which is simply $\sqrt{2}$ using this corrected definition of $z$. The setup for bounds on the integral are still OK, although since now we're integrating a constant we don't need to do the integral, which will of course turn out to be the area of the circle in the $xy$ plane, multiplied by $\sqrt{2}$.

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