Question

Compute the surface area of revolution about the x-axis over the interval [0, 1] for $y = e^{-2x}$. (Use symbolic notation and fractions where needed.) $S = -\frac{1}{2} \pi \int_0^1 2x \left( \left( \frac{1}{2} \sqrt{1 + (4e^{-4x})^2} (4e^{-4x}) + \frac{1}{2} \ln \left( \left| \sqrt{1 + (4e^{-4x})^2} + 4e^{-4x} \right| \right) \right) - \left( \frac{1}{2} \sqrt{1 + (4e^0)^2} (4e^0) + \frac{1}{2} \ln \left( \left| \sqrt{1 + (4e^0)^2} + 4e^0 \right| \right) \right) \right) dx$

          Compute the surface area of revolution about the x-axis over the interval [0, 1] for $y = e^{-2x}$.
(Use symbolic notation and fractions where needed.)
$S = -\frac{1}{2} \pi \int_0^1 2x \left( \left( \frac{1}{2} \sqrt{1 + (4e^{-4x})^2} (4e^{-4x}) + \frac{1}{2} \ln \left( \left| \sqrt{1 + (4e^{-4x})^2} + 4e^{-4x} \right| \right) \right) - \left( \frac{1}{2} \sqrt{1 + (4e^0)^2} (4e^0) + \frac{1}{2} \ln \left( \left| \sqrt{1 + (4e^0)^2} + 4e^0 \right| \right) \right) \right) dx$

$Compute the surface area of revolution about the x-axis over the interval [0, 1] for y = e^-2x. (Use symbolic notation and fractions where needed.) S = -(1)/(2)π∫0^1 2x ( ( (1)/(2)√(1 + (4e^-4x)^2) (4e^-4x) + (1)/(2)ln( | √(1 + (4e^-4x)^2) + 4e^-4x| ) ) - ( (1)/(2)√(1 + (4e^0)^2) (4e^0) + (1)/(2)ln( | √(1 + (4e^0)^2) + 4e^0 | ) ) ) dx$

Added by Scott S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Mukesh Devi

05/08/2022

Step 1

Step 1: To compute the surface area of revolution about the x-axis, we need to use the formula: \[S = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\] where $y$ is the function, $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$, Show more…

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THIS IS INCORRECT:527/125PLEASE PUT INTO SYMBOLIC NOTATION AND/OR FRACTIONS. Compute the surface area of revolution about the x-axis over the interval 01for y=e2 (Use symbolic notation and fractions where needed.)