# Finding the Area of S: A=a(S)

• Dafe
In summary, the conversation involves finding the area of a region S in the xy-plane defined by the equations z=x+y^2, 0<=x<=1, and x<=y<=1. The correct representation of the region is either 0<=x<=1, x<=y<=1 or 0<=y<=1, 0<=x<=y.

## Homework Statement

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

## Homework Equations

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

## 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) = $$\int_0^1\int_y^1 \sqrt{2+4y^2}\,dy\,dx$$

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

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

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.