How do you find the surface area of a boundary in $\mathbb{R}^3$? I need to solve this problem:

Let $D=\{(x,y,z):4(x-2+z)^2+4y^2\le(2-z)^2,0\le x-z\le1\}$
Calculate the area of $\partial D$

So how do you calculate the area of the boundary of a volume defined like this?
I googled till my eyes started bleeding, but since I'm italian I fear I may be using the wrong technical terms for my searches.
Can anybody help me? Thanks.
 A: In the first part,we will show what the surface of $D$ looks alike.
We start with $D=\{(x,y,z):4(x-2+z)^2+4y^2\le(2-z)^2,0\le x-z\le1\}......(1)$.
Since $0\le x-z\le1$ we can set $x-z =t$ and 
$$0\le t\le1......(2)$$ 
Substituting $x=z+t$ into (1), we can solve for $y$ and obtain:
$$y_1(t,z) \le y \le y_2(t,z)=-y_1(t,z)......(3)$$
where
$$y_1(t,z)=-(1/2)\left((3 z-2 + 2 t ) (6 - 2 t - 5 z)\right)^{1/2}......(4)$$
$$y_2(t,z)=(1/2)\left((3 z-2 + 2 t ) (6 - 2 t - 5 z)\right)^{1/2}......(5)$$
Solving $y_1(t,z)=0$ we obtain that 
$$z_1(t):=(2/3)(1-t) \le z \le (2/5)(3-t)=:z_2(t)......(6)$$
Thus we may define $z=z_1(t) (1-s)+z_2(t) s$ and $0 \le s \le 1$
And
$$y_1(t,s)=-(2/\sqrt{15})(2+t)\left(s(1-s)\right)^{1/2}......(7)$$
$$y_2(t,s)=(2/\sqrt{15})(2+t)\left(s(1-s)\right)^{1/2}......(8)$$
The plot of the $\partial D$ (minus the front and back ellipses is depicted in Figure below:
In the second part, we will calculate the area of $\partial D$.
Define 
$$\vec r(t,s) = \{x(t,s),y(t,s),z(t,s)\}=\left\{t+z_1(t)(1-s)+z_2(t)s,y_2(t,s),z_1(t)(1-s)+z_2(t)s)\right\}$$
The area of the top part of $\partial D$ is then given by:
$$A_1=\int_0^1 \int_0^1|\vec r_t \times \vec r_s| dt ds=\int_0^1\frac{2+t}{15}dt \int_0^1 \left(\frac{212s^2-220s+125}{15s(1-s)}\right)^{1/2}ds$$
$$=\frac16 \int_0^1 \left(\frac{212s^2-220s+125}{15s(1-s)}\right)^{1/2}ds$$
This can be expressed as elliptical integral.
The area of the bottom part of $\partial D$ is then given by $A_2=A_1$.
The area of the back part is bounded by curves $y_1(0,s)=-(4/\sqrt{15})\sqrt{s(1-s)}$ and $y_2(0,s)=(4/\sqrt{15})\sqrt{s(1-s)}$ and is given by
$$A_3=\frac{\pi}{15}$$
The area of the front part is bounded by curves $y_1(1,s)=-(12/5)^{1/2}\sqrt{s(1-s)}$ and $y_2(1,s)=(12/5)^{1/2}\sqrt{s(1-s)}$ and is given by
$$A_4=(3/10)^{1/2}\pi$$
