Area of the region $\{(x,y):0\leq x \leq 1, 0 \leq y\leq 1, 3/4\leq x+y\leq 3/2\}$

Question

Find the area of the region $\{(x,y):0\leq x \leq 1, 0 \leq y\leq 1, 3/4\leq x+y\leq 3/2\}$ (using definite integration).

I cannot understand how to find this area. I have graphed the lines and found out the required region. I found the definite integral $\int_{0}^{1} (3/2-y)-(3/4-y)dy$ but it is yielding an extra areas. How do I find the area of the region?

Answer 1

Since $y \leq 1$, it is :

$$\frac{3}{4} \leq x + y \leq \frac{3}{2} \Rightarrow -\frac{1}{4} \leq x \leq \frac{1}{2}$$

But $x \geq 0$, thus :

$$0 \leq x \leq \frac{1}{2}$$

Then, for $y$ one would get :

$$\frac{1}{4} \leq y \leq 1$$

The desired area of $D = \{(x,y) \in \mathbb R^2 : x \geq 1, y \leq 1, \frac{3}{4} \leq x+y \leq \frac{3}{2}\}$, is :

$$A(D) = \iint_D \mathrm{d}x\mathrm{d}y = \int_0^\frac{1}{2} \int_{\frac{1}{4}}^1\mathrm{d}x\mathrm{d}y$$

Answer 2

We should be able to find the area of this polygon without calculus.

At the very least, you should be able to divide it up into a bunch of triangles.

by the shoelace algorithm I get:

$\begin {array}{} \frac 34 & 0\\ 1&0\\ 1&\frac 12\\ \frac 12& 1\\ 0&1\\ 0&\frac 34 \end{array}$

$\frac 12( \frac 12 + 1 + \frac 12 - \frac 14 - \frac 9{16}) = \frac {19}{32}$

If you want to use calculus, most direct would be

$\int_0^{\frac 12} 1 - (\frac 34 - x) \ dx + \int_{\frac 12}^{\frac 34} (3/2 - x) - (\frac 34 - x) \ dx + \int_{\frac 34}^1 3/2 - x \ dx$

Answer 3

Using the symmetry of given construction, we can consider it in a coordinate system, rotated by $45^\circ$

Then

\begin{align} [ABECDF]&= 2\left( \int_{\tfrac{3\sqrt2}8}^{\tfrac{\sqrt2}2} x\,dx +\int_{\tfrac{\sqrt2}2}^{\tfrac{3\sqrt2}4} (\sqrt2-x) \, dx \right) =\frac{19}{32} . \end{align}

Answer 4

As other answers have noted, the region is a square with two isosceles right-angles triangles cut off, making its area$$1-\tfrac12\left(\left(\tfrac12\right)^2+\left(\tfrac34\right)^2\right)=1-\tfrac12\cdot\tfrac{13}{16}=\tfrac{19}{32}.$$