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Surface Area of Revolution: Calculus Exercises and Solutions, Assignments of Differential and Integral Calculus

A series of exercises and solutions related to the concept of surface area of revolution in calculus. It includes detailed steps for solving problems involving the rotation of curves around the x-axis and y-axis. Valuable for students studying calculus, particularly those focusing on integral calculus and its applications.

Typology: Assignments

2021/2022

Uploaded on 01/03/2025

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Find the surface area surface generated when y=2x is revolved
about the y-axis from x=0 to x=4
SOLUTION:
y = 2x
Surface Area of Revolution
DIFFERENTIAL AND INTEGRAL CALCULUS, Feliciano and Uy, Chapter
12.10, page 365, no. 11
R
y=3-
2x+x2
Y =
3
ESM 1031 ENGINEERING CALCULUS 2 (INTEGRAL CALCULUS)
Surface area of Revolution
MODULE 2 – DEFINITE INTEGRALS
[RENDE MAE PACLIBAR – 1H COURSE]
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Find the surface area surface generated when y=2x is revolved about the y-axis from x=0 to x= SOLUTION: y = 2x

Surface Area of Revolution

DIFFERENTIAL AND INTEGRAL CALCULUS, Feliciano and Uy, Chapter 12.10, page 365, no. 11 R y=3- 2x+x^2

Y =

ESM 1031 ENGINEERING CALCULUS 2 (INTEGRAL CALCULUS)

Surface area of Revolution

MODULE 2 – DEFINITE INTEGRALS

[RENDE MAE PACLIBAR – 1H COURSE]

Find the surface area surface generated when y=2x is revolved about the y- axis from x=0 to x= CONTINUATION… S = S = S =

Surface Area of Revolution

DIFFERENTIAL AND INTEGRAL CALCULUS, Feliciano and Uy, Chapter 12.10, page 365, no. 11 R y=3- 2x+x^2

Y =

ESM 1031 ENGINEERING CALCULUS 2 (INTEGRAL CALCULUS)

Surface area of Revolution

MODULE 2 – DEFINITE INTEGRALS

[RENDE MAE PACLIBAR – 1H COURSE]

Find the area of the surface generated when arc y= is revolved about the x-axis from x=0 to x=1. Given: Req’d: Area of the surface generated SOLUTION:

Surface Area of Revolution

DIFFERENTIAL AND INTEGRAL CALCULUS, Feliciano and Uy, Chapter 12.10, page 365, no. 2 y= x= ESM 1031 ENGINEERING CALCULUS 2 (INTEGRAL CALCULUS) Surface area of Revolution

MODULE 2 – DEFINITE INTEGRALS

[RENDE MAE PACLIBAR – 1H BSCE]

= x= S =

3 2