Question

Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated line.

About the line y = - 6

x = y + 6
x=y^2

Answer 1

Just a moment. Only just noticed this.

Answer 2

y+6=y^2 then y=-2 and 3 . then 2pie integral -2 to 3 (y^2)(y+6)
my answer is 345pi/2 but it must be 378pi/2

Answer 3

Sorry. I'm gettin' there.

Answer 4

Integral(-2 to 3) {2pi(y-(-6))*(y+6 -y^2)}
Tell me if you see that.

Answer 5

Limits of integration are right. Then you need 2pi*r. The radius is y-(-6) or y+6. Then you need the width of the tube, which is the difference of the two functions.

Answer 6

sorry, where did you get this? what is the formula?

Answer 7

Alright, that's wrong. But I'm not quite sure how. Gimme a second.

Answer 8

ok
thanks

Answer 9

Baallls. Something is wrong. I feel so sure of my answer, but it doesn't match what's given.

Answer 10

It seems to me, you consider a particular y, right? And for that y, you calculate two things. The circumference of a tube around the axis is the first. The length of that tube is the second.

Answer 11

because the answer he gave was wrong...you are correct smoothmath

Answer 12

what is the formula when we have to curved and shell method? 2pi integral f(x)g(x)?

Answer 13

Well the first answer I gave was wrong.

Answer 14

http://www.wolframalpha.com/input/?i=integrate+%28y%2B6-y^2%29*PI+*2%28y%2B6%29+dy+from+y%3D-2+to+3
I think that's correct.

Answer 15

could u plz just help me with the way?

Answer 16

\[\int\limits_{-2}^{3}2\pi(y+6)(y+6-y^2)dy\]

is correct

Answer 17

Refer to picture.

Answer 18

Okay, you see that yellow line? That's the difference between the two functions. Computationally, it's (y+6-y^2).

Answer 19

your line is incorrect

Answer 20

The idea is that you rotate that line in a circle around the axis y=-6 and you calculate the volume it creates. The volume is the length of that line multiplied by the diameter of the resulting circle.

Answer 21

the x=y+6

y=x-6

Answer 22

Make any sense fifi?

Answer 23

you could also use disks

and get

\[\int\limits_{0}^{4}\pi((\sqrt{x}+6)^2-(-\sqrt{x}+6)^2)dx+\int\limits_{4}^{9}\pi((\sqrt{x}+6)^2-(x-6+6)^2)dx\]

Answer 24

no. i don't have problem with risk. i have problem with shell. specially when i have two curves. i do not know if i have to deduct, multiple or what?

Answer 25

I only did it that way to verify my answer...both ways give me the same answer

Answer 26

what is your answer?
and what is shell way? could u plz write it?

Answer 27

\[\frac{1625\pi}{6}\]

Answer 28

\[\int\limits_{-2}^{3}2\pi(y+6)(y+6-y^2)dy\]

Answer 29

Fifi, to do shell, think of making a lot of hollow tubes. So that yellow line in my picture gets rotated around the axis and forms one tube. It's one of an infinite number of tubes. So then you add up all those tubes by integrating and it gives you the volume.

Answer 30

sorry, i want to know when we have f(x) and g(x) how we can find volume by shell? please do help me. i have lots of questions and i have to take my exam before 8pm

Answer 31

∫−232π(y+6)(y+6−y2)dy how do u find it what is the formula?

Answer 32

2piintegral f(x)(f(x)-g(x))
??

Answer 33

\[\int\limits_{a}^{b}2\pi(x-r)(f(x)-g(x))dx\]

here r is below the enclosed area and f(x) is above g(x)

Answer 34

if r is above the enclosed area then we have

\[\int\limits_{a}^{b}2\pi(r-x)(f(x)-g(x))dx\]

or you can just put in the absolute value \[|x-r|\]

Answer 35

going to gym..good luck