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Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]

Ask by Crawford Mcdonald. in the United States
Mar 15,2025

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The surface area of the rotated curve \( y = x^3 \) over \( 0 \leq x \leq 2 \) is \[ S = \frac{\pi}{27} \left(145^{\frac{3}{2}} - 1\right). \]

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Bonus Knowledge

Ah, the joy of using calculus to explore the surfaces of revolution! The concept of finding the surface area of a curve can be traced back to the works of pioneers like Archimedes, who first tackled the volumes and areas of revolution. His ancient methods laid the groundwork for what would eventually evolve into the integral calculus we use today. So, next time you're calculating surface areas, you’re standing on the shoulders of giants who literally shaped the math world! Now, let’s talk about how this applies in real life! Engineers and architects often need to calculate the surface area of various components, whether it's designing elegant shapes for bridges or analyzing the material needed for complex structures like water tanks. Rotating curves can give rise to all sorts of practical applications in fields such as 3D printing and manufacturing where efficiency and accuracy are crucial. So, the next time you’re admiring a beautifully designed building or product, you might just be witnessing the integration of calculus in action!

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