How Do I Sketch the Region for Finding the Area of S?

[Dafe]

Homework Statement

S={(x,y,z): z=x+y^2, 0\leq x\leq1 , x\leq y \leq 1

Homework Equations

A=a(S) = $<br /> \int\int_R \sqrt{(1+(\frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2\,dy\,dx

The Attempt at a Solution

S={(x,y,z): z=x+y^2, 0\leq x\leq1 , x\leq y \leq 1

I suppose I can write this as:

S={(x,y,z): z=x+y^2, y\leq x\leq1 , 0\leq y \leq 1
And so i think:

A=a(S) = $<br /> \int_0^1\int_y^1 \sqrt{2+4y^2}\,dy\,dx

If I calculate this I don't get the answer that I should..

 

[Dick]

It would help a great deal to know what you got and how you got it.

 

[arildno]

SKETCH the region given!

Do it in the following way:

1. Draw the rectangle strip 0\leq{x}\leq{1} in the xy-plane.
2. Since y<=1, draw the line y=1. You are to be below this line, within the strip from 1.

3. Now, you have x<=y. Draw the line x=y, you are to be above that line.

4. Thus, you may represent the region as follows:
0\leq{x}\leq{1}, x\leq{y}\leq{1}
These limits on y were gained by looking at the vertical line segments the region consists of for all x-positions of these segments from 0 to 1.

Alternatively, we may consider the horizontal line segments the region consists of; this yields the equally valid representation:
0\leq{y}\leq{1}, 0\leq{x}\leq{y}


These are the two simplest correct region representations, yours is not correct.