A surface in $\mathbb{R}^3$ with equation $z = xy$ intersects a surface with equation $y = 2x^2$ along some curve. Each point of the arc of this curve contained in the area $0 < z < 1$ is connected by a line segment to the point $(0,0,0)$. The sum of these segments (without ends) forms the surface of $M$. Determine that $M$ is a manifold and calculate the area of the surface of $M$.
Graph of $z = xy$ in $\mathbb{R}^3$ looks like that:
Graph of $y = 2x^2$ in $\mathbb{R}^2$ looks like that:
So we can imagine that some curve on $z = xy$ plane that is the reflection of $y = 2x^2$ curve that lays in the plane underneath. We can now imagine that we connect each point of that curve with the point (0,0,0) and need to calculate the surface of the plane that we obtain by doing so.
In the intersection we know that:
- $y = 2x^2$
- $z = xy \implies z = 2x^3$
We know that: $z \in (0,1) \implies x \in \left( 0,\left( \frac{1}{2} \right)^{\frac{1}{3}} \right)$
I would like to integrate over $x$ and $t \in (0,1)$, where:
- for $t = 0$ we would be in point $(0,0,0)$
- for $t = 1$ we would be in points on curve
Therefore, we get such a parametrization:
- $x \to tx$
- $y \to 2tx^2$
- $z \to 2tx^3$
$$\begin{pmatrix} x & t \\ 2x^2 & 4tx \\ 2x^3 & 6tx^2 \end{pmatrix}$$
From that I can calculate jacobian:
$$\begin{pmatrix} x & t \\ 2x^2 & 4tx \\ 2x^3 & 6tx^2 \end{pmatrix}^T \cdot \begin{pmatrix} x & t \\ 2x^2 & 4tx \\ 2x^3 & 6tx^2 \end{pmatrix} = $$
$$= \begin{pmatrix} x & 2x^2 & 2x^3 \\ t & 4tx & 6tx^2 \end{pmatrix} \cdot \begin{pmatrix} x & t \\ 2x^2 & 4tx \\ 2x^3 & 6tx^2 \end{pmatrix} = $$
$$= \begin{pmatrix} x^2 + 4x^4 + 4x^6 & xt +8tx^3 + 12tx^5 \\ xt +8tx^3 + 12tx^5 & t^2 + 16t^2x^2 + 36t^2 x^4 \end{pmatrix}$$
From that I calculate the determinant:
$$(x^2 + 4x^4 + 4x^6)(t^2 + 16t^2x^2 + 36t^2 x^4) - (xt +8tx^3 + 12tx^5)^2 =$$
$$= (x^2 + 4x^4 + 4x^6)(t^2 + 16t^2x^2 + 36t^2 x^4) - (xt + 8tx^3 + 12tx^5)(xt + 8tx^3 + 12tx^5) =$$
$$= x^2t^2 + 16t^2x^4 + 36t^2x^6 + 4t^2x^4 + 64t^2x^6 + 144t^2 x^8 + 4t^2x^6 + 64t^2x^8 + 144t^2x^{10} - x^2t^2 - 8t^2x^4 - 12t^2x^6 - 8t^2x^4 - 64t^2x^6 - 96t^2x^8 - 12t^2x^6 - 96t^2x^8 - 144t^2x^{10} =$$
$$= 4t^2x^4 + 16t^2x^6 + 16t^2x^8 = 4t^2x^4 (1 + 4x^2 + 4x^4)$$
Then we know that square root of that is equal to $\sqrt{4t^2x^4 + 16t^2x^6 + 16t^2x^8}$
From that we calculate the integral:
$$\int_0^1 \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} \sqrt{4t^2x^4 + 16t^2x^6 + 16t^2x^8} \ dx \ dt = $$
$$= 4 \int_0^1 t \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} \sqrt{x^4(1 + 4x^2 + 4x^4)} \ dx \ dt = $$
$$= 4 \int_0^1 t \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} \sqrt{x^4(1 + 2x^2)^2} \ dx \ dt = $$
$$= 4 \int_0^1 t \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} x^2(1 + 2x^2) \ dx \ dt = $$
$$= 4 \int_0^1 t \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} x^2 + 2x^4 \ dx \ dt = $$
$$= 4 \int_0^1 t \left( \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} x^2 \ dx + \int_0^{\left( \frac{1}{2} \right)^{\frac{1}{3}}} 2x^4 \ dx \right) \ dt = $$
$$= 4 \int_0^1 t \left(\frac{1}{6} + \frac{2}{5} \left( \frac{1}{2} \right)^{\frac{5}{3}} \right) \ dt = $$
$$= 4 \left(\frac{1}{6} + \frac{2}{5} \left( \frac{1}{2} \right)^{\frac{5}{3}} \right) \int_0^1 t \ dt = $$
$$= 4 \left(\frac{1}{6} + \frac{2}{5} \left( \frac{1}{2} \right)^{\frac{5}{3}} \right) \frac{1}{2} = $$
$$= 2 \left(\frac{1}{6} + \frac{2}{5} \left( \frac{1}{2} \right)^{\frac{5}{3}} \right)$$
Does it look correct?