Given the area is in the first quadrant bounded by y²=x, the line x=4 and the 0X
What is the volume generated when this area is revolved about the 0X?
the answer is 25.13 but I don't know how. help me please
dV=2PI y(dA)
where dA=dy*(4-y^2)) from y=0 to 2
dV=2PI(4y-y^3)dy from y= 0 to 2
V= 2PI ( 2y^2-y^4/4)
V=2PI (8-4)=8PI=your answer
or using discs
V π∫ y^2 dx from x= 0 to 4
= π∫x dx from x = 0 to 4
= π[ (1/2)x^2] from 0 to 4
= π(16/2 - 0)
= 8π
Well, when it comes to revolve things, it's all about getting into the spin of things! In this case, we have a region in the first quadrant that's bounded by the curve y² = x, the line x = 4, and the x-axis. We want to spin this beautiful area around the x-axis and calculate the resulting volume.
To find the volume generated, we can use the method of cylindrical shells or the disk method. Let's go with the disk method because it's less likely to end up like a circus act gone wrong.
The first thing we need to do is find the boundaries of the region. We have the line x = 4 and the curve y² = x. To find where they intersect, we can substitute x = 4 into the equation of the curve:
y² = 4
y = ±2
So our region is bounded by the x-axis, the curve y² = x, and the line x = 4. Great!
Now, let's slice this region into very thin disks, like those tiny delicious pancakes. Each disk will have a thickness dx and a radius determined by the value of x for each slice.
The formula for the volume of each disk is V = π(radius)²(thickness). In this case, the radius is just y, which is the square root of x, and the thickness is dx.
So, the volume of each disk is V = π(x)(dx).
Now, we just need to integrate this expression from x = 0 to x = 4 to sum up the volumes of all the disks:
∫[0 to 4] π(x)(dx)
Evaluating this integral gives us:
π(x²/2) from 0 to 4
Plugging in the boundary values, we get:
π(4²/2) - π(0²/2)
π(8) - π(0)
8π - 0
8π
And there you have it! The volume generated when this area is revolved about the x-axis is 8π.
Now, if we want to calculate the numerical value, we can use the approximation π ≈ 3.14:
8π ≈ 8 * 3.14 ≈ 25.12
So, the volume generated is approximately 25.12 (or 25.13, if we want to round it off).
Hope that clears things up and brings a smile to your face, even if math can sometimes be a circus!
To find the volume generated when the given area is revolved about the 0X axis, we can use the method of cylindrical shells.
First, let's sketch the given region in the first quadrant. The region is bounded by the parabola y^2 = x, the line x = 4, and the 0X-axis.
Since the region is symmetric about the 0X-axis, we can focus on the portion in the first quadrant. The region can be visualized as a quarter of a circle with a quadrant of a parabola removed.
To find the volume, we need to integrate the area of the cylindrical shells that make up the solid formed by revolving the region about the 0X-axis.
The radius of each cylindrical shell will be the x-coordinate of a point on the boundary of the region (y² = x), and the height of each cylindrical shell will be the length of the curved boundary segment that is revolved.
Let's set up the integral:
V = ∫(2πrh) dx,
where r is the radius and h is the height of each cylindrical shell.
Knowing that y² = x, we can solve for y:
y = √x.
The lower and upper limits of integration will be the x-values at which the region starts and ends.
The region starts at x = 0 (0X-axis) and ends at x = 4 (boundary line x = 4).
Using the equation of a parabola, we can substitute y = √x into the equation and solve for x:
√x² = x.
This implies that the region begins and ends at x = 0, so the integral limits are from x = 0 to x = 4.
Calculating the height of each cylindrical shell is simply the difference in y-values at a particular x:
h = y - 0 = √x - 0 = √x.
Now, let's calculate the radius of each cylindrical shell:
r = x.
Substituting the values of r and h into the integral, we have:
V = ∫(2πxr)(√x) dx
= ∫(2πx^(3/2)) dx.
Integrating with respect to x:
V = (2π) * (∫x^(3/2) dx)
= (2π) * (2/5 * x^(5/2))|_0^4
= (2π) * (2/5 * (4)^(5/2) - 0^(5/2))
= (2π) * (2/5 * 8 * √4)
= (2π) * (2/5 * 8 * 2)
= (2π) * (32/5)
= 64π/5
≈ 12.73.
So, the volume generated when this area is revolved about the 0X-axis is approximately 12.73 cubic units.
It seems that the answer you were given (25.13) might be incorrect. Please double-check the given information or let me know if there are any additional details that need to be considered.
To find the volume generated when a region is revolved about an axis, you can use the method of cylindrical shells or the method of washers.
In this case, let's use the method of cylindrical shells:
1. Sketch the region bounded by the curve y² = x, the line x = 4, and the x-axis. This region is a parabolic shape in the first quadrant.
|
y |
| | x=4
|____|___________ <- Curve y² = x
|
|____________
| x-axis
2. To generate the volume, imagine taking thin horizontal slices of the region and rotating them around the x-axis. Each slice will form a cylindrical shell.
3. Consider a small horizontal slice at a distance x from the origin. The height of this slice will be the difference between the upper and lower y-values of the region at that x-coordinate. From the equation y² = x, we can solve for y: y = √x.
4. The radius of the cylindrical shell will be the distance from the x-axis to the curve. Since the curve is y² = x, the radius will be y.
5. The thickness (dx) of the cylindrical shell will be a small change in x.
6. Express the volume of each cylindrical shell as V = 2πrhdx, where r is the radius, h is the height, and dx is the thickness.
7. Integrate the volume formula over the range of x-values that form the region. In this case, integrate from x = 0 to x = 4.
V = ∫(from 0 to 4) 2πy√xdx
8. Substituting y = √x into the volume formula, we have:
V = 2π∫(from 0 to 4) (√x)(√x)dx
= 2π∫(from 0 to 4) x dx
= 2π[x²/2] (from 0 to 4)
= 2π(4²/2 - 0²/2)
= 16π
9. Calculate the value of 16π, which is approximately 50.27.
Therefore, the volume generated when the given region is revolved about the x-axis is approximately 50.27 cubic units.