Find the surface area bounded by (x+y)²+z=1, x>0, y>0, z>0
Find the surface area bounded by $(x+y)^2+z=1$, $x>0,\,y>0,\,z>0$.
Explanation of Method
The solid is bounded above by the paraboloid $z=1-(x+y)^2$ over the triangle $x,y>0,\,x+y<1$, below by $z=0$ (the same triangle), and on the sides by the planes $x=0$ and $y=0$. Four separate pieces to compute and sum: the curved paraboloid patch (via the change of variables $u=x+y,\,v=x-y$), the flat triangular base, and two flat parabolic-segment side faces.
- Compute the paraboloid patch $S_1$ via $u=x+y,\,v=x-y$.
- Compute the base triangle's area $S_2$.
- Compute the two side faces $S_3,S_4$ and sum all four.
Step 1 · Paraboloid patch $S_1$
With $z=1-(x+y)^2$: $z_x=z_y=-2(x+y)$, so $1+z_x^2+z_y^2=1+8(x+y)^2$. With $u=x+y,\,v=x-y$ (Jacobian $\frac12$), the triangle becomes $0<u<1,\,-u<v<u$:
$$ S_1 = \int_0^1\int_{-u}^u\sqrt{1+8u^2}\cdot\frac12\,dv\,du = \int_0^1u\sqrt{1+8u^2}\,du = \frac1{24}\big(27-1\big) = \frac{13}{12}. $$
Step 2 · Base triangle $S_2$
The projection is the triangle with vertices $(0,0),(1,0),(0,1)$: $S_2=\frac12$.
Step 3 · Side faces $S_3,S_4$
On $x=0$: $z=1-y^2$ for $0<y<1$, a parabolic-segment region:
$$ S_3 = \int_0^1(1-y^2)\,dy = \frac23, $$
and by symmetry $S_4=\frac23$ (on $y=0$).
Step 4 · Total
$$ S = S_1+S_2+S_3+S_4 = \frac{13}{12}+\frac6{12}+\frac8{12}+\frac8{12} = \frac{35}{12}. $$
Additional Notes
- This derivation matches the source exactly, including correctly identifying and summing all four bounding faces of the solid.
- Verified numerically: $S_1\approx1.0833$, $S_2=0.5$, $S_3=S_4\approx0.6667$, total $\approx2.9167$, matching $\dfrac{35}{12}$ exactly.
Result
The total surface area is: