Surface Area Let be a nonnegative function such that is continuous over the interval Let

Chapter 15, Problem 58

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Area Let f be a nonnegative function such that f ‘ is continuous over the interval [a, b]. Let S be the surface of revolution formed by revolving the graph of f, where \(a \leq x \leq b\), about the x-axis. Let x = u, y = f(u) cos v, and z = f(u) sin v, where \(a \leq u \leq b\) and \(0 \leq v \leq 2 \pi\). Then, S is represented parametrically by \(\mathbf{r}(u, v)=u \mathbf{i}+f(u) \cos v \mathbf{j} +f(u) \sin v \mathbf{k}\). Show that the following formulas are equivalent.

Surface area \(=2 \pi \int_{a}^{b} f(x) \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x\)

Surface area \(=\int_{D} \int\left\|\mathbf{r}_{u} \times \mathbf{r}_{v}\right\| d A\)

Text Transcription:

a leq x leq b

a leq u leq b

0 leq v leq 2

r(u, v) = ui + f(u) cos vj  + f(u) sin vk

= 2 pi int_{a}^{b} f(x) sqrt{1 + [f ‘(x)]^2} dx

= int_D int || r_u X r_v || dA