Sketch the region enclosed by the curves x=64−y^2 and x=y^2−64. Decide whether to integrate with respect to x or y. Then find the area of the region.
Area =
I assume you sketched the two curves.
You should have two horizontal parabolas that have the x-axis as their axes of symmetry, and y-intercepts of (0, ±8) and x-intercepts of (±64,0)
best way: Due to the nice symmetry, find the area of the region in the first quadrant and multiply that by 4, by using horizontal slices.
Area = 4 ∫ x dy from 0 to 8
= 4∫ (64-y^2) dy from 0 to 8
= 4 [64y - y^3/3] from 0 to 8
= 4(512 - 512/3 - (0-0) )
= 4(1024/3)
= 4096/3
check my arithmetic
Well, well, well, we have ourselves a little enclosed region here! Let's see what we've got.
We have two curves, x = 64 - y^2 and x = y^2 - 64. Now, let's think about how these curves are shaped.
If we look at the first curve, x = 64 - y^2, we see that it's a nice little parabola opening downwards. And if we take a glance at the second curve, x = y^2 - 64, we see it's another parabola, but this time it opens upwards.
So, what does this mean for finding the area? Well, if we want to find the area of the region enclosed by these curves, we need to integrate with respect to y. Why, you ask? Because the curves are defined in terms of y, silly!
So, we're going to integrate with respect to y. Now, let's find the limits of integration. To do that, we need to find the y-values where the two curves intersect. We can set the two equations equal to each other:
64 - y^2 = y^2 - 64
Simplify that bad boy and you'll get 2y^2 = 128. Dividing both sides by 2 gives us y^2 = 64, and taking the square root of both sides leaves us with y = ±8.
So we have our limits of integration: y = -8 to y = 8.
Now, it's time to find the area using good old integration. Let's integrate the difference of the two curves from y = -8 to y = 8:
Area = ∫[from -8 to 8] (y^2 - 64) - (64 - y^2) dy
Now, let's do some fancy math and solve that integral. Integrate each term separately and you'll find:
Area = ∫[from -8 to 8] 2y^2 - 128 dy
Integrating that, you'll get:
Area = [2/3 * y^3 - 128y] from -8 to 8
Plugging in the limits, you'll get:
Area = [2/3 * (8^3) - 128(8)] - [2/3 * (-8^3) - 128(-8)]
Simplify that expression and you'll find:
Area = 1024 - (-1024)
The negatives cancel out, and we're left with:
Area = 2048 square units.
So, there you have it! The area of the region enclosed by those curves is 2048 square units. Ta-da!
To find the area of the region enclosed by the curves x = 64−y^2 and x = y^2−64, we need to first sketch the region.
Let's start by graphing the curves.
The curve x = 64−y^2 is a downward-opening parabola with its vertex at (0, -64) and its axis of symmetry along the y-axis.
The curve x = y^2−64 is an upward-opening parabola with its vertex at (0, 64) and its axis of symmetry along the y-axis.
Now, let's find the points of intersection between the two curves.
Setting the x values equal to each other, we have:
64 - y^2 = y^2 - 64
Rearranging the equation:
2y^2 = 128
Dividing both sides by 2:
y^2 = 64
Taking the square root of both sides:
y = ±8
So, the curves intersect at the points (8, 8) and (-8, -8).
Next, we need to determine whether to integrate with respect to x or y. Since the curves are defined in terms of x in terms of y, we will integrate with respect to y.
The limits of integration for y are from -8 to 8.
Therefore, the area of the region can be calculated by integrating the difference between the two curves with respect to y:
Area = ∫[from -8 to 8] (x = 64−y^2) - (x = y^2−64) dy
Area = ∫[from -8 to 8] (64−y^2) - (y^2−64) dy
Simplifying the equation, we get:
Area = ∫[from -8 to 8] 128 - 2y^2 dy
Now we can integrate:
Area = [128y - (2/3)y^3] evaluated from -8 to 8
Substituting the limits of integration:
Area = [128(8) - (2/3)(8)^3] - [128(-8) - (2/3)(-8)^3]
Area = [1024 - (2/3)(512)] - [-1024 - (2/3)(-512)]
Area = 1024 - (1024/3) + 1024 + (1024/3)
Simplifying the equation, we get:
Area = 2048/3
So, the area of the region enclosed by the curves x = 64−y^2 and x = y^2−64 is 2048/3.
To find the area of the region enclosed by the curves x = 64 - y^2 and x = y^2 - 64, we first need to sketch the curves.
Let's start with the curve x = 64 - y^2. This is a downward-facing parabola with its vertex at (0, -64). It opens towards the negative x-axis.
Next, let's consider the curve x = y^2 - 64. This is an upward-facing parabola with its vertex at (0, -64). It opens towards the positive x-axis.
To determine the region enclosed by these curves, we want to find the intersection points where the two curves intersect.
Setting the equations equal to each other, we have:
64 - y^2 = y^2 - 64
Rearranging the equation, we get:
y^2 = 64/2 = 32
Taking the square root of both sides, we find:
y = ± √32 = ± 4√2
Therefore, the y-values of the intersection points are y = 4√2 and y = -4√2.
Now that we have identified the intersection points, we can determine the boundaries for integration.
Since the curves intersect at y = 4√2 and y = -4√2, we integrate from y = -4√2 to y = 4√2.
To find the area, we integrate with respect to y.
Using the formula for finding the area between two curves, we have:
Area = ∫[from y = -4√2 to y = 4√2] [top curve: (64 - y^2) - (y^2 - 64)] dy
Simplifying the equation, we have:
Area = ∫[from y = -4√2 to y = 4√2] (128 - 2y^2) dy
Evaluating the integral, we get:
Area = [128y - (2/3)y^3] |_y=-4√2 ^y=4√2
Plugging in the values, we have:
Area = (128 * 4√2 - (2/3)(4√2)^3) - (128 * -4√2 - (2/3)(-4√2)^3)
Simplifying further, we get:
Area = (512√2 - (2/3)(32√2)) - (-512√2 - (2/3)(32√2))
Area = 512√2 - 64√2 + 512√2 + 64√2
Area = 1088√2
Therefore, the area of the region enclosed by the curves x = 64 - y^2 and x = y^2 - 64 is 1088√2, when integrating with respect to y.