91Ó°ÊÓ

Integral calculus is all about finding the total effect of something by summing up small pieces. In this exercise, we're focusing on the *Surface Area of Revolution*. Here, we use integrals to calculate the surface area formed when a curve is revolved around an axis. When revolving a curve, the formula for surface area involves an integral, because the calculation needs to account for the continuously changing shape of the curve. For revolution around the y-axis, as in this exercise, the formula is: - \[ S = \int 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] Breaking this down:- The \(2\pi x\) represents the circumference of a small slice parallel to the y-axis.- \(dS\), the arc differential, adds the complexity of the curve into our calculation.Understanding how these pieces come together in an integral is key to solving many real-world problems involving curves, like finding the area of a surface, the volume of a solid, or the length of a curve.

Arc length is the distance along a curve between two points. In our problem, we're using the arc length to help determine the surface area of revolution. The differential arc length, \(dS\), covers the small length of a curve segment. When you add up these small segments, you find the total length of the arc. The formula for arc length combines both the x component and the y component:- \[dS = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \cdot dx\] This expression provides a way to factor in how your curve changes as both x and y change. The \(\frac{dy}{dx}\) is the slope of the curve at any point x, giving an insight into how steep or flat a segment of the curve is. When working with curves in calculus, remember:- Calculate \(\frac{dy}{dx}\) accurately as it heavily impacts your final calculations.- Integrating \(dS\) over an interval gives you the overall arc length, vital for accurately determining the characteristics of a curve's surface area.

The chain rule is an essential tool in calculus, especially for handling problems involving composite functions. It shines through in our problem as we calculated the derivative \(\frac{dy}{dx}\). Understanding it well helps simplify complicated functions into easier parts. To use the chain rule, you must recognize when a function is composed of other functions. In our exercise, the function \(y = \frac{1}{3}(x^2 + 2)^{3/2}\) was the composite of the inner function \(u = x^2 + 2\) and the outer function \(f(u) = u^{3/2}\).Here's how the chain rule applies:- **Identify the inner and outer functions**: - Inner: \(u = x^2 + 2\) - Outer: \(f(u) = u^{3/2}\)- **Differentiate the outer function with respect to the inner function** and the inner function with respect to \(x\): - \(\frac{df}{du} = \frac{3}{2}u^{1/2}\) - \(\frac{du}{dx} = 2x\)- **Apply Chain Rule**: - \(\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{3}{2}u^{1/2} \cdot 2x = x(x^2 + 2)^{1/2}\)Using the chain rule effectively allows a problem that initially looks daunting to become straightforward and solvable. Mastery of it unlocks otherwise difficult calculus problems.