Question

The given curve is rotated about the x-axis. Set up, but do not evaluate, an integral for the area of the resulting surface by integrating (a) with respect to x and (b) with respect to y. $$x = \ln(8y + 1), \quad 0 \le y \le 1$$ (a) Integrate with respect to x. $$\int_{0}^{\ln(9)} \left( \left( \frac{e^x - 1}{8} \right) \sqrt{1 + \left( \frac{e^x}{8} \right)} \right) dx$$ (b) Integrate with respect to y. $$\int_{0}^{1} \left( \ln(8y + 1) \sqrt{1 + \left( \frac{8}{8y + 1} \right)^2} \right) dy$$

          The given curve is rotated about the x-axis. Set up, but do not evaluate, an integral for the area of the resulting surface by integrating (a) with respect to x and (b) with respect to y.
$$x = \ln(8y + 1), \quad 0 \le y \le 1$$
(a) Integrate with respect to x.
$$\int_{0}^{\ln(9)} \left( \left( \frac{e^x - 1}{8} \right) \sqrt{1 + \left( \frac{e^x}{8} \right)} \right) dx$$
(b) Integrate with respect to y.
$$\int_{0}^{1} \left( \ln(8y + 1) \sqrt{1 + \left( \frac{8}{8y + 1} \right)^2} \right) dy$$

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the given curve is rotated about the x axis set up but do not evaluate an integral for the area of the resulting surface by integrating a with respect to x and b with respect to y x ln8y 1 q 28504

Added by Daniel R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Jason Gerber

11/17/2022

Step 1

We need to do this in two ways: (a) integrating with respect to x, and (b) integrating with respect to y. The formula for the surface area of revolution about the x-axis is given by: $S = \int 2\pi y \, ds$ where $ds = \sqrt{1 + (y')^2} \, dx$ if integrating with Show more…

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The given curve is rotated about the x-axis. Set up, but do not evaluate, an integral for the area of the resulting surface by integrating (a) with respect to x and (b) with respect to y. $$x = \ln(8y + 1), \quad 0 \le y \le 1$$ (a) Integrate with respect to x. $$\int_{0}^{\ln(9)} \left( \left( \frac{e^x - 1}{8} \right) \sqrt{1 + \left( \frac{e^x}{8} \right)} \right) dx$$ (b) Integrate with respect to y. $$\int_{0}^{1} \left( \ln(8y + 1) \sqrt{1 + \left( \frac{8}{8y + 1} \right)^2} \right) dy$$

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