How to determine the area of the paraboloid enclosed by the cone?

Question

Is it possible to determine the exact area of the paraboloid that falls inside the cone?
I've been trying for days without success...

Answer 1

HINT: This is a surface integral: $$\iint_A 1\ ds=\iint_T\sqrt{\left( \frac{\partial f(x,y)}{\partial x}\right)^2+\left( \frac{\partial f(x,y)}{\partial y}\right)^2+1}\ dxdy,$$ where $$f(x,y)=\frac{x^2+y^2}{4a},\ \frac{\partial f(x,y)}{\partial y}=\frac{y}{2a},\ \frac{\partial f(x,y)}{\partial x}=\frac{x}{2a}$$ the paraboloid and its partial derivatives and $$T=\{x,y:z=\frac{x^2+y^2}{4a}\text{ and }x^2+(z-a)^2=\left(\frac{R_s}{r_{\theta}}\right)^2y^2\}=$$ $$=\{x,y:x^2+(\sqrt{\frac{x^2+y^2}{4a}}-a)^2=\left(\frac{R_s}{r_{\theta}}\right)^2y^2\},$$ that is, the shadow of the patch ($A$) on the surface formed by the paraboloid and the cone.

Having said all that "all you need to do" is to evaluate the little integral below:

$$\iint_T\sqrt{\left( \frac{x}{2a}\right)^2+\left( \frac{y}{2a}\right)^2+1}\ dxdy.$$

The toughest part is to reorganize $T$ so the integration could be actually performed...

Regarding $T$: one can realize that the equation describing the shadow can be written in the following form:

$$ex^4+fx^2+gx^2y^2+hy^2+i^4+j=0,$$

where the constants depend on $a, \text{ and } \theta$. The good news is that $x$ does not appear in the equation. Now, substitute $u=x^2$. The result is a nice second order equation in $u$:

$$eu^2+u(f+gy^2)+hy^2+iy^4+j=0.$$

If you consider $y$ to be a constant then you can solve this equation for $u$, and as a result, for $x$. The last step before integrating is to see the range of $y$ (We already know the lower bound $y_1=r_{\theta}\ $.) So, you are going to have an integral of the form:

$$\int_{y_1}^{y_2}\left(\int_{x_1(y)}^{x_2(y)}\cdots \ \ dx\right)dy,$$

where $x_1(y)$ and $x_2(y)=...$ are the solutions of the equation above. This my be more complicated than I described, but it is certainly doable.

Now, somebody will have to work hard...