how do you find the volume of the region defined by y = sqrt(x), y = 0 and x = 6 when the region is revolving around the line x = 6?

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

you can do areas of circles then....
pi {S} [sqrt(x)]^2 dx ; [0,6]

- amistre64

pi x^2
------ at 6 = 18pi
2

- amistre64

sooo close... i forgot to read the hwhole thing

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

since you are at the line x=6; subtract 6 from the function to bring it down to x = 0

- amistre64

over down..someplace; same idea

- amistre64

sqrt(x+6) is the function you use to move everything to the x = 0 line

- anonymous

okay....so my 18 pi answer was wrong; where do you subtract 6 from? I did and equation where it was just sqrt(x) and then one where it was sqrt(x)-6, and for the first one I found it to be 18pi, and the second came out to be a weird decimal

- amistre64

##### 1 Attachment

- anonymous

that's the graph I have, yep

- amistre64

let me redo what I said lol

- anonymous

okay

- amistre64

pi {S} [sqrt(x+6)]^2 dx ; [0,6]
pi {S} x+6 dx ; [0,6]
pi(x^2/2 + 6x) at x = 6

- anonymous

you added 6 under the sqrt sign. How do you know to do that?

- amistre64

thats how we move things left and right.... see for yourself; when you put in x = -6 you want a zero right? to match the original

- anonymous

OKAY. wow, brain lapse.

- amistre64

we dont want to cahng ethe value of the function; just move it :)

- amistre64

and I steered wrong again i think; I sppun it around the x axis in that..

- anonymous

so you used the disk method to solve the problem

- amistre64

you can use disk; but youd end up doing it with respect to y and would have to change the formulas again.. shell is just as well

- amistre64

when I just did it I spun it around the x axis like an idiot lol

- amistre64

2pi {S} x(sqrt(x-6)) dx is good too

- anonymous

using shell method would leave me with \[2\pi \int\limits_{0}^{6}(6-x)(\sqrt(x+6)dx ?\]

- amistre64

maybe harder; so lets reinterpret the graph for y instead of x... does that make sense?

- anonymous

let's try

- amistre64

not 6-x; just x; the radius is x as x moves from 0 to 6, but that integral looks scary

- anonymous

yes it does

- amistre64

sqrt(x^2(x+6)) = sqrt(x^3 + 6x^2) and integrating that looks painful lol

- anonymous

\[\pi \int\limits_{0}^{6}y(y^2+6)dy ?\]

- amistre64

x = y^2-6 is a good tranlation dont you think?

- anonymous

oh, y^2 minus 6. But the rest of the integration looks okay?

- amistre64

hmm.... if we shell it with respect to y, that doesnt get us out spin around the x=0 part

- amistre64

yes; but we need a new interval; one that works along the y line

- anonymous

now i am confused...can we try again with x? That makes the most sense to me

- amistre64

when x = 6 to begin with; y = sqrt(6) so our interval is along the y axis from 0 to sqrt(6)

- amistre64

if we do it with respect to x; we run into an awful integral.....

- anonymous

okay, now I understand. The limits were wrong

- amistre64

if we do it with respect to y; its easier, we just have to know how far to go onthe y to match

- amistre64

at x= 6, our original; y = sqrt(6) so our boundary is from 0 to sqrt(6) of y^2-6

- anonymous

and it's minus 6 because of "top-bottom"?

- anonymous

\[\pi \int\limits_{0}^{\sqrt(6)}y(y^2-6)dy\]

- amistre64

##### 1 Attachment

- amistre64

now we do the disc method of areas of circles..... to spin around x=0

- anonymous

yeah! now I get it

- amistre64

pi {S} [y^2-6]^2 dy; from [0,sqrt(6)]

- anonymous

wait, so my integral is wrong again, since we revert to disk method?

- amistre64

correct; remember shell method gave us a nightmare integral

- amistre64

we might get a negative out of this; but that just means we went the wrong direction on the interval and just ignore the sign

- amistre64

pi {S} (y^4 -12y^2 +36) dy ints to:
pi(y^5/5 -12y^3/3 + 36y) at y = sqrt(6) right?

- anonymous

yeah; okay so now my answer is -9.797958971, the same answer I had, but didn't get right. Is there a fractional way of writing this?

- anonymous

my bad. recalculating...

- amistre64

((sqrt(6)^5) / 5) - ((12 * (sqrt(6)^3)) / 3) + (36 * sqrt(6)) = 47.0302031 according to the google

- anonymous

hmmm

- anonymous

I put that into my calculator and got -940898/7275....

- amistre64

http://www.wolframalpha.com/input/?i=int%28pi%28y^2-6%29^2%29+dy+from+0+to+sqrt%286%29

- amistre64

i mighta typoed it in google lol

- anonymous

no, you have yours right I think. What the fractional form?

- anonymous

it's right!!! thanks a bunch!

- anonymous

I have another problem just like that so I will use this method

- amistre64

:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.