Use the shell method to find the volume of the solid generated by revolving the

Question

Use the shell method to find the volume of the solid generated by revolving the region bounded by the line (y=5x+6) and the parabola (y=x^2) about the following lines: a) x=6, b) x=-1, c) the x axis, and d) y=36.

Please keep in mind that I care about HOW to do this problem. Please be explicit because I have no clue how to use the shell method when it is not around the axis. Here are the answers to the first two parts. Please show me how to get them and how I choose what is the shell height and what is the shell radius.

a = 2401pi/6, b= 2401pi/6

Explanation / Answer

a)
Using shell method:

height --------> 5x + 6 - x^2
radius ====> 6 - x

limits:
5x + 6 = x^2
0 = x^2 - 5x - 6
0 = (x - 6)(x + 1) ----> x = -1 & 6

6
? 2? * (6 - x) * ( 5x + 6 - x^2 ) dx = 2401?/6
-1

-----------
b)

Using shell method:

height --------> 5x + 6 - x^2
radius ====> x + 1

6
? 2? * (x + 1) * ( 5x + 6 - x^2 ) dx = 2401?/6
-1

-----------
c)
Using shell method:

y = x^2 ----> x = +/- ?(y)
y = 5x + 6 ---> x = (y - 6)/5

height --------> ?(y) - -?(y) -----> 2?(y)
radius --------> y
when x = -1 ------> y = 5(-1) + 6 = 1

height --------> ?(y) - (y - 6)/5
radius --------> y
when x = 6 ------> y = 5(6) + 6 = 36

1. . . .. . . . . . . . . . . . .36
? 2? * y * ( 2?(y) ) dy + ? 2? * y * ( ?(y) - (y - 6)/5 ) dy = 23324?/15
0. . . .. . . . . . . . . . . . .1

--------
d)

Using shell method:

1. . . .. . . . . . . . . . . . . . . . .36
? 2? * (36 - y) * ( 2?(y) ) dy + ? 2? * (36 - y) * ( ?(y) - (y - 6)/5 ) dy = 38416?/15
0. . . .. . . . . . . . . . . . . . . . . 1

========free to e-mail if need steps

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.