Find the exact area of the surface obtained by rotating the given curve about the x-axis.
x=t^3, y=t^2, 0 ≤ t ≤ 1
?[0,1] 2?x ds
= ?[0,1] 2?x?(x'^2 + y'^2) dt
= ?[0,1] 2?t^3?((3t^2)^2 + (2t)^2) dt
Not an easy one -- it will involve integration by parts.
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,1%5D+(2%CF%80t%5E3%E2%88%9A((3t%5E2)%5E2+%2B+(2t)%5E2))+dt
Oh, we're doing math, aren't we? Don't worry, I won't clown around too much with the answer. To find the surface area obtained by rotating the curve about the x-axis, we can use the formula for surface area of revolution.
The formula is: A = 2π ∫[a, b] y √(1 + (dy/dx)^2) dx
Now let's put on our serious face and do some calculations. The curve is given by x = t^3 and y = t^2, where 0 ≤ t ≤ 1.
First, let's find dy/dx:
dy/dx = (dy/dt) / (dx/dt)
= (2t) / (3t^2)
= 2/(3t)
Now let's calculate the integral:
A = 2π ∫[0, 1] t^2 √(1 + (2/(3t))^2) dt
Ugh, this integral is really getting under my clownish skin. It's a bit messy, but nothing too scary. So, let me just crunch the numbers and get back to you.
To find the exact surface area obtained by rotating the curve about the x-axis, we can use the formula for the surface area of a curve generated by rotating about the x-axis, which is given by:
\[A = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\dfrac{dx}{dt}\right)^2} dt \]
In this case, the given curve is x = t^3 and y = t^2, with a = 0 and b = 1.
First, let's find \(\dfrac{dx}{dt}\):
\[\dfrac{dx}{dt} = \dfrac{d(t^3)}{dt} = 3t^2 \]
Now, let's substitute the given values into the surface area formula:
\[A = \int_{0}^{1} 2\pi t^2 \sqrt{1 + (3t^2)^2} dt \]
Simplifying the square root:
\[A = \int_{0}^{1} 2\pi t^2 \sqrt{1 + 9t^4} dt \]
Integrating this expression is quite complicated, and there is no elementary antiderivative for this function. Therefore, we need to use numerical methods or approximations to find the surface area.
To find the exact area of the surface obtained by rotating the given curve about the x-axis, we can use the formula for the surface area of revolution:
A = ∫ 2πy √(1 + (dy/dx)^2) dx
First, let's find dy/dx by taking the derivative of y with respect to x:
dx/dt = 3t^2
dy/dt = 2t
dy/dx = (dy/dt) / (dx/dt) = (2t) / (3t^2) = 2/3t
Now we can rewrite the formula for the surface area:
A = ∫ 2πy √(1 + (dy/dx)^2) dx
= ∫ 2πt^2 √(1 + (2/3t)^2) (3t^2) dx
= ∫ 2πt^2 √(1 + (4/9t^2)) (3t^2) dx
= ∫ 2πt^2 √((9t^2 + 4) / 9t^2) (3t^2) dx
= ∫ 2πt^2 (3t) √(9t^2 + 4) dx
Now, we need to determine the limits of integration. The given range for t is 0 ≤ t ≤ 1. We are rotating the curve about the x-axis, so the limits of integration for x will correspond to the values of x at those t-values.
When t = 0, x = 0^3 = 0
When t = 1, x = 1^3 = 1
Therefore, the limits of integration are 0 to 1.
Now we can proceed with the integration:
A = ∫ 2πt^2 (3t) √(9t^2 + 4) dx
= 2π ∫ t^3 √(9t^2 + 4) dx
= 2π ∫ √(9t^2 + 4) t^3 dx
Unfortunately, we have reached a point where we cannot evaluate the integral analytically. We would need to use numerical methods, such as numerical integration or approximation techniques, to find an approximate value for the area.
Alternatively, if you have a specific value of t or a particular range of t in mind, I can help you find the numerical approximation for the area within that range.