Question

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the $y$-axis.\ $y = \frac{x}{2} + 6$, $1 \le x \le 9$\ $2\pi \int_1^9 \left( \boxed{\qquad} \right) dx = 24\pi$

Name: Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the y-axis.y = (x)/(2) + 6, 1 ≤ x ≤ 92π∫1^9 ( ) dx = 24π
Uploaded: 2024-06-06T20:42:35-08:00
Duration: 5 min
Channel: Zhumagali Shomanov
Description: Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the y-axis.y = (x)/(2) + 6, 1 ≤ x ≤ 92π∫1^9 ( ) dx = 24π

          Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the $y$-axis.\
$y = \frac{x}{2} + 6$, $1 \le x \le 9$\
$2\pi \int_1^9 \left( \boxed{\qquad} \right) dx = 24\pi$

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the y-axis.y = (x)/(2) + 6, 1 ≤ x ≤ 92π∫1^9 ( ) dx = 24π

Added by Aaron Z.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

06/06/2024

Step 1

Step 1: The formula for the surface area of a solid of revolution is given by: $$S = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} dx$$ where $f(x)$ is the function that defines the curve, and $a$ and $b$ are the limits of integration. Show more…

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Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the y-axis. $$y = \frac{x}{2} + 6,$$ 1 ≤ x ≤ 9 $$2\pi \int_1^9 ( \boxed{ } ) dx = 24\pi$$