Question

Sketch the region in the x y-plane defined by the inequalities x - 2y²≥0,1 - x - | y | ≥ 0 and find its area.

Short answer

The area of the shaded region is \(\frac{7}{{12}}\)

Step-by-step solution

Solve the curve function

Solve for x

x - 2y²≥0

x ≥ 2y²

x = 2y²

(Boundary line)

To graph this, draw the boundary line (change inequality to equal)\(x = 2{y^2}\) as a solid line (since the inequality is "...or equal to"). Then the shaded part is to the left of the boundary because it is x greater than ...".

Pick points to test, if needed more confirmation.

Solve the other one for \(x\)

1 - x - | y| ≥ 0

1 - | y | ≥ x

x ≤ 1 - | y |

To get rid of the absolute value split into two parts

x ≤ 1 - y, if y ≥ 0

x ≤ 1 - (-y), if y ≤ 0

The boundary lines (solid lines) will be these as equations

\(x = 1 - y\)

and \(x = 1 + y\), and the shaded area is to the left since the inequality is " \(x\) less than ...".

Plot the graph from the obtained function.

The shaded areas is the enclosed region.

Find the intersections

To find the intersections, set the boundary equations equal to each other. Since they are symmetric about the \(x\)-axis use the \(x = 1 - y\) part and neglect the other to get the \(y\)-coordinate.

2y² = 1 - y

2y² + y - 1 = 0

(2y - 1)(y + 1) = 0

\(y = \frac{1}{2}, - 1\)

Ignore the \(y = - 1\) one because that would be if \(x = 1 - y\) were extended past \((1,0)\). So the \(y\)-coordinate of the other intersection would be \(y = - \frac{1}{2}\).

Find the area of the region.

Also because of the symmetry, integrate from \(y = 0\) to \(y = \frac{1}{2}\), and multiply that by 2.Integrate the top function with respect to \(y\) is the one more to the right, which would be \(x = 1 - y\).

\(\begin{aligned}A &= 2\int_0^{1/2} {\left( {1 - y - 2{y^2}} \right)} dy\\ &= 2\left[ {y - \frac{1}{2}{y^2} - \frac{2}{3}{y^3}} \right]_0^{1/2}\\ &= 2\left[ {\frac{1}{2} - \frac{1}{2}{{\left( {\frac{1}{2}} \right)}^2} - \frac{2}{3}{{\left( {\frac{1}{2}} \right)}^3} - (0 - 0 - 0)} \right]\\ &= 2\left[ {\frac{1}{2} - \frac{1}{8} - \frac{2}{{24}}} \right]\end{aligned}\)

\(\begin{aligned} &= 2\left[ {\frac{{12 - 3 - 2}}{{24}}} \right]\\ &= \frac{7}{{12}}\end{aligned}\)