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?##