Find the area enclosed by the equations x=16-y^2 and x=y^2-16
I don't think the equations enclose any area. But their graphs do. Gotta be precise when talking and doing math!
The curves intersect at y=±4, so the area between the curves is just
∫[-4,4] (16-y^2)-(y^2-16) dy
which, taking advantage of the symmetry is
4∫[0,4] 16-y^2 dy
= 4(16y - 1/3 y^3) [0,4]
= 4(64 - 64/3)
= 512/3
You could also take vertical strips instead of horizontal, so the area is
4∫[0,16] √(16-x) dx
= 8/3 (16-x)^(3/2)
= 512/3
To find the area enclosed by the equations x=16-y^2 and x=y^2-16, we first need to graph the equations and determine the points of intersection.
To do this, we can start by rearranging both equations to solve for y in terms of x.
For the equation x=16-y^2, we can subtract x from both sides to get -y^2 = -x + 16, and then multiply by -1 to get y^2 = x - 16. Taking the square root, we have y = ±√(x - 16).
Next, for the equation x=y^2-16, we can add 16 to both sides to get x + 16 = y^2, and then take the square root to get ±√(x + 16) = y.
Now we can plot the graphs of these two equations on a coordinate plane.
First, let's consider the equation y = √(x - 16). Since the square root of a number is always positive, this equation represents the upper half of the graph. Similarly, the equation y = -√(x - 16) represents the lower half of the graph.
Next, let's consider the equation y = √(x + 16). Similarly, this equation represents the upper half of the graph, and y = -√(x + 16) represents the lower half.
Now, plotting the graphs of these equations on a coordinate plane, we can see that they intersect at two points: (-16, 0) and (16, 0).
To find the area enclosed by the curves, we need to integrate the difference between the upper and lower curves with respect to x, and then find the absolute value of the result. Since there are two curves and each has a different range, we need to calculate two separate integrals.
Let's start with the part of the curve between the points of intersection (-16, 0) and (16, 0). We integrate the upper curve (y = √(x + 16)) minus the lower curve (y = √(x - 16)) with respect to x:
∫[from -16 to 16] (√(x + 16) - √(x - 16)) dx
Simplifying this integral and evaluating it will give us the area of this portion.
Next, we consider the parts of the curves on either side of the point of intersection (16, 0). For these parts, we integrate the upper curve (y = √(x - 16)) minus the lower curve (y = -√(x + 16)) with respect to x:
∫[from 16 to a value greater than 16] (√(x - 16) - (-√(x + 16))) dx
Again, simplifying and evaluating this integral will give us the area of this portion.
Finally, we add the areas of these two portions together to find the total enclosed area.
Please note that the exact numerical values of these areas depend on the limits of integration and the specific values of x. The general approach explained above is applicable to finding the area enclosed by any set of equations, but the specific calculations may vary depending on the specific equations and their properties.