Find the area bounded by the y-axis and x = 4-y^2/3
I will assume you meant:
x = 4 - y^(2/3)
I assume you made a sketch
let's take horizontal slices ....
the y-intercept is (0,8) so the region is not closed.
I will assume you want the region between the x-axis and and the y-axis
area = ∫4 - y^(2/3) dy from 0 to 8
= 4y - (3/5)y^(5/3) | from 0 to 8
= (32 - (3/5)(8^(5/3)) - 0
= 32 - (3/5)(32)
= 64/5 or 12.8
check my arithmetic
Well, if you want to find the area bounded by the y-axis and the curve x = 4 - y^2/3, you can think of it as finding the area of a quirky-looking shape. To proceed, we first need to figure out the limits of integration.
Let's solve the equation x = 4 - y^2/3 for y. By rearranging the equation, we get y^2/3 = 4 - x, so y^2 = 3(4 - x), and finally, y = ±√(3(4 - x)).
Now, to find the limits of integration, we need to determine where these two curves intersect. Since we're looking for the area bounded by the y-axis, x = 0, we can substitute x = 0 into y = ±√(3(4 - x)). This gives us y = ±√12.
So, our limits of integration are y = -√12 to y = √12.
To find the area, we need to integrate the curve x = 4 - y^2/3 with respect to y over the given limits.
Unfortunately, things get a bit complicated here, and my sense of humor can't come up with a simple explanation. But don't worry, I'm always here to put a smile on your face with a joke or two!
To find the area bounded by the y-axis and the curve x = 4 - y^(2/3), we need to find the points of intersection between the curve and the y-axis and then calculate the area between these points.
To find the points of intersection, we set x = 0 in the equation of the curve:
0 = 4 - y^(2/3)
Rearranging the equation to solve for y:
y^(2/3) = 4
Taking the cube of both sides:
y^2 = 64
Taking the square root of both sides:
y = ±8
So, the two points of intersection between the curve and the y-axis are (0, 8) and (0, -8).
To calculate the area between these points, we integrate the curve from y = -8 to y = 8:
∫[0,8] (4 - y^(2/3)) dy
To integrate this expression, we can first simplify the equation inside the integral:
∫[0,8] (4 - y^(2/3)) dy = 4y - (3/5)y^(5/3) evaluated from 0 to 8
Evaluating the antiderivative at the upper limit (8) and subtracting the value at the lower limit (0):
= (4(8) - (3/5)(8)^(5/3)) - (4(0) - (3/5)(0)^(5/3))
Simplifying further:
= (32 - (3/5)(8)^(5/3)) - 0
= 32 - (3/5)(8)^(5/3)
Calculating the value:
≈ 32 - (3/5)(8.717)
≈ 32 - (3/5)(8.717)
≈ 32 - (3/5)(14.670)
≈ 32 - 8.802
≈ 23.198
Therefore, the area bounded by the y-axis and the curve x = 4 - y^(2/3) is approximately 23.198 square units.
To find the area bounded by the y-axis and the curve defined by the equation x = 4 - y^(2/3), we can use integration.
The first step is to determine the limits of integration. Since the curve is defined by the equation x = 4 - y^(2/3), we can rearrange it to solve for y: y^(2/3) = 4 - x.
Taking both sides to the power of 3/2, we get y^2 = (4 - x)^(3/2).
To find the limits of integration for y, we need to solve this equation for y. When y^2 = 0, we have y = 0. When (4 - x)^(3/2) = 0, we have 4 - x = 0, which means x = 4.
Therefore, our limits of integration for y are from y = 0 to y = (4 - x)^(3/2).
Now, to find the area bounded by the y-axis and the curve, we integrate the function with respect to y from 0 to (4 - x)^(3/2):
A = ∫[0, (4 - x)^(3/2)] 1 dy.
This integral represents the area between the curve and the y-axis. To evaluate it, we integrate 1 with respect to y:
A = [y] evaluated from 0 to (4 - x)^(3/2).
Substituting the limits of integration, we get:
A = (4 - x)^(3/2) - 0,
A = (4 - x)^(3/2).
Hence, the area bounded by the y-axis and x = 4 - y^(2/3) is given by (4 - x)^(3/2).