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In order to find the area, first draw the inequality.
72π square units.
We want to find the area of the region represented by the following inequality. x^2+y^2 ≤ 72 In order to do so, let's first find the region given by it.
Graphing an inequality involves two main steps.
We can tell a lot of information about the boundary curve from the given inequality.
Let's find each of these key pieces of information for the inequalities in the system.
Information | Inequality |
---|---|
Given Inequality | x^2+y^2≤72 |
Boundary Curve Equation | x^2+y^2=72 |
Solid or Dashed? | ≥ ⇒ Solid |
To graph the boundary curve equation, we will make a table of values.
x | x^2+y^2=72 | y |
---|---|---|
- 6 | ( - 6)^2+y^2=72 | ± 6 |
- 3 | ( - 4)^2+y^2=72 | ± sqrt(63) ≈ ± 7.94 |
0 | ( 0)^2+y^2=72 | ± sqrt(72) ≈ ± 8.49 |
3 | ( 3)^2+y^2=72 | ± sqrt(63) ≈ ± 7.94 |
6 | ( 6)^2+y^2=72 | ± 6 |
We will now plot and connect the obtained points with a smooth curve.
As we can see, this curve is a circle with (0,0) as a center and a radius of sqrt(72) or about 8.49 units.
Before we can shade the solution set for the inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary curve.
x= 0, y= 0
Calculate power
Add terms
r= sqrt(72)
sqrt(a^2)=a
Multiply