Question

6.6.17 Find the area of the surface generated when the given curve is revolved about the x-axis. y = ?(-x^2 - 2x + 8) on [?3, ?2] The area of the surface is ____ square units. (Type an exact answer, using ? as needed.)

          6.6.17
Find the area of the surface generated when the given curve is revolved about the x-axis.
y = ?(-x^2 - 2x + 8) on [?3, ?2]
The area of the surface is ____ square units.
(Type an exact answer, using ? as needed.)

Added by James W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/21/2023

Step 1

The formula for the surface area of revolution about the x-axis is given by: $$A = 2\pi\int_a^b y\sqrt{1+(\frac{dy}{dx})^2} dx$$ In this case, the curve is given by $y = 2x + 8$, and we want to find the surface area on the interval $[-3, -2]$. Show more…

Show all steps

Thanks for your feedback!

6.6.17 Find the area of the surface generated when the given curve is revolved about the x-axis. y = x^2 + 8 on [-3, -2]. The area of the surface is ____ square units. (Type an exact answer, using π as needed.) Enter your answer in the answer box and then click Check Answer.