If you do not answer the above questions, you will not have a correct answer.

[oteggis]

Homework Statement

Determine the area of the surface A of that portion of the paraboloid:
[x][/2]+[y][/2] -2z = 0

where [x][/2]+[y][/2]≤ 8 and y≥x

Homework Equations

Area A = ∫∫ dS

The Attempt at a Solution

Area A = ∫∫ dS = 3∫∫ dS

 

[Charles Link]

The boundary of the projected area onto x-y plane has $x^2+y^2 \leq 8$ , but that doesn't mean in evaluating $\sqrt{x^2+y^2+1}$ that $x^2+y^2=8$.

 

[oteggis]

√x2+y2+1x2+y2+1)dA gives dS

 

[Charles Link]

You need to compute $S=\iint \sqrt{x^2+y^2+1} \, dxdy$ over the specified region of covered by $x$ and $y$ in the x-y plane. Also see post 2 again.

 

[oteggis]

You need to compute $S=\iint \sqrt{x^2+y^2+1} \, dxdy$ over the specified region of covered by $x$ and $y$ in the x-y plane. Also see post 2 again.

This will give Area = ∫∫ 3rdrdθ using cylindrical coordinates

 

[Charles Link]

Why are you assuming $x^2+y^2=8$? $\\$ $x$ and $y$ need to be integrated over the circle where $r<\sqrt{8}=2 \sqrt{2}$ and they cut that region in half by saying that $y>x$. $\\$ If you want to use polar coordinates $x^2+y^2=r^2$, and $r$ must be integrated from $0$ to $2 \sqrt{2}$. You will have $\sqrt{r^2+1 }$ as the integrand.

 

[Ray Vickson]

This will give Area = ∫∫ 3rdrdθ using cylindrical coordinates

No, it will not.

Can you have x=0 and y=0 for a point on the sueface? In other words, using the surface equation, is there a point z(x,y) = z(0,0) on it? can you have x=1/2 and y= 2 on the surface? That is, do we have a point z(x,y) = z(1/2,2) on it? do (x,y) = (0,0) and (x,y) = (1/2,2) satisfy the equation $x^2+y^2 = 8$? Do they satisfy the inequality $x^2+y^2 \leq 8?$