Find the exact area of the surface obtained by rotating the curve about the x-axis.
y=((x^3)/4)+(1/3X), 1/2≤X≤1
just plug and chug. The surface can be thought of as a stack of thin rings.
A = ∫2πr ds
where r=y and ds=√(1+y'^2) dx
so,
y' = (9x^4-4)/(12x^2)
A = 2π∫[1/2,1] (x^3/4+1/(3x))*((9x^4+4)/(12x^2)) dx = 1981π/3072
Oh, rotating curves! I like to think of it as giving the curve a little spin. You know, like a fancy dance move. But instead of a dance floor, we're using the x-axis.
So to find the area of the surface obtained by rotating the curve y = (x^3)/4 + (1/3)x about the x-axis, we'll need to integrate.
But wait, before we dive into the math, let's find the interval of integration. You've already given us that 1/2 ≤ x ≤ 1. That's like telling our dance move to stay within a particular part of the dance floor. Nice and limited.
Alright, let's get back to it. To find the surface area, we can use the formula:
A = 2π ∫[a,b] y * √(1 + (dy/dx)^2) dx.
Don't worry, I won't leave you hanging with this math stuff. I'm here to help!
Now, let's find the derivative dy/dx of the curve y = (x^3)/4 + (1/3)x. The derivative is dy/dx = (3x^2)/4 + 1/3.
Next, plug this derivative into the formula and integrate over the given interval [1/2, 1].
A = 2π ∫[1/2, 1] ((x^3)/4 + (1/3)x) * √(1 + ((3x^2)/4 + 1/3)^2) dx.
And there you have it! Dance your way through that integral, and you'll find the exact surface area. Remember, math can be fun, just like dancing!
To find the surface area obtained by rotating the curve about the x-axis, we can use the formula for a surface of revolution:
A = 2π ∫[a, b] y * sqrt(1 + (dy/dx)^2) dx
In this case, the given curve is y = (x^3)/4 + (1/3)x and the interval is 1/2 ≤ x ≤ 1.
To begin, let's find dy/dx:
dy/dx = (3/4)x^2 + (1/3)
Now, let's substitute the curve equation and the derivative into the surface area formula:
A = 2π ∫[1/2, 1] [(x^3)/4 + (1/3)x] * sqrt(1 + [(3/4)x^2 + (1/3)]^2) dx
Simplifying the expression under the square root:
A = 2π ∫[1/2, 1] [(x^3)/4 + (1/3)x] * sqrt(1 + (9/16)x^4 + (2/3)x^2 + (1/9)) dx
We can distribute the integration over the sum:
A = 2π ∫[1/2, 1] (x^3/4) * sqrt(1 + (9/16)x^4 + (2/3)x^2 + (1/9)) dx
+ 2π ∫[1/2, 1] (1/3)x * sqrt(1 + (9/16)x^4 + (2/3)x^2 + (1/9)) dx
The integrals on both terms are complicated, and there is no closed-form solution for the antiderivative of the square root expression. Therefore, we need to approximate the integral using numerical methods or a computer software.
To find the exact area of the surface obtained by rotating the curve about the x-axis, we can use the method of integration known as the disk method.
1. First, let's sketch the curve and the region of rotation. The given equation is y = (x^3)/4 + (1/3)x. The range of x-values is from 1/2 to 1. When we rotate this curve about the x-axis, it forms a solid with a circular cross-section.
2. Next, let's express the equation in terms of x as a function of y. Rearrange the equation to solve for x: x = (4y - (1/3)y^2)^(1/3)
3. Now, let's consider an infinitesimally thin slice of the solid at a particular y-value. This slice will have a radius equal to the x-value at that y-location.
4. To find the radius of the disk, we substitute the function of x in terms of y into the equation and simplify: r = (4y - (1/3)y^2)^(1/3)
5. The area of an infinitesimally thin disk is given by the formula A = πr^2. Substitute the expression for r found in step 4 to get A = π[(4y - (1/3)y^2)^(1/3)]^2
6. To find the total area of the rotation, we integrate this expression with respect to y over the given y-range (1/2 to 1). The integral will be ∫[1/2, 1] π[(4y - (1/3)y^2)^(1/3)]^2 dy
7. Evaluate the integral to find the exact area of the surface obtained by rotating the curve about the x-axis.
Note: Since this is a mathematical calculation which involves complex steps, it may be easier to use math software or an online integrator to evaluate the integral and obtain an exact numerical value for the area.