anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
you can do areas of circles then.... pi {S} [sqrt(x)]^2 dx ; [0,6]
amistre64
  • amistre64
pi x^2 ------ at 6 = 18pi 2
amistre64
  • 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
  • amistre64
since you are at the line x=6; subtract 6 from the function to bring it down to x = 0
amistre64
  • amistre64
over down..someplace; same idea
amistre64
  • amistre64
sqrt(x+6) is the function you use to move everything to the x = 0 line
anonymous
  • 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
  • amistre64
1 Attachment
anonymous
  • anonymous
that's the graph I have, yep
amistre64
  • amistre64
let me redo what I said lol
anonymous
  • anonymous
okay
amistre64
  • 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
  • anonymous
you added 6 under the sqrt sign. How do you know to do that?
amistre64
  • 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
  • anonymous
OKAY. wow, brain lapse.
amistre64
  • amistre64
we dont want to cahng ethe value of the function; just move it :)
amistre64
  • amistre64
and I steered wrong again i think; I sppun it around the x axis in that..
anonymous
  • anonymous
so you used the disk method to solve the problem
amistre64
  • 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
  • amistre64
when I just did it I spun it around the x axis like an idiot lol
amistre64
  • amistre64
2pi {S} x(sqrt(x-6)) dx is good too
anonymous
  • anonymous
using shell method would leave me with \[2\pi \int\limits_{0}^{6}(6-x)(\sqrt(x+6)dx ?\]
amistre64
  • amistre64
maybe harder; so lets reinterpret the graph for y instead of x... does that make sense?
anonymous
  • anonymous
let's try
amistre64
  • amistre64
not 6-x; just x; the radius is x as x moves from 0 to 6, but that integral looks scary
anonymous
  • anonymous
yes it does
amistre64
  • amistre64
sqrt(x^2(x+6)) = sqrt(x^3 + 6x^2) and integrating that looks painful lol
anonymous
  • anonymous
\[\pi \int\limits_{0}^{6}y(y^2+6)dy ?\]
amistre64
  • amistre64
x = y^2-6 is a good tranlation dont you think?
anonymous
  • anonymous
oh, y^2 minus 6. But the rest of the integration looks okay?
amistre64
  • amistre64
hmm.... if we shell it with respect to y, that doesnt get us out spin around the x=0 part
amistre64
  • amistre64
yes; but we need a new interval; one that works along the y line
anonymous
  • anonymous
now i am confused...can we try again with x? That makes the most sense to me
amistre64
  • amistre64
when x = 6 to begin with; y = sqrt(6) so our interval is along the y axis from 0 to sqrt(6)
amistre64
  • amistre64
if we do it with respect to x; we run into an awful integral.....
anonymous
  • anonymous
okay, now I understand. The limits were wrong
amistre64
  • 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
  • amistre64
at x= 6, our original; y = sqrt(6) so our boundary is from 0 to sqrt(6) of y^2-6
anonymous
  • anonymous
and it's minus 6 because of "top-bottom"?
anonymous
  • anonymous
\[\pi \int\limits_{0}^{\sqrt(6)}y(y^2-6)dy\]
amistre64
  • amistre64
1 Attachment
amistre64
  • amistre64
now we do the disc method of areas of circles..... to spin around x=0
anonymous
  • anonymous
yeah! now I get it
amistre64
  • amistre64
pi {S} [y^2-6]^2 dy; from [0,sqrt(6)]
anonymous
  • anonymous
wait, so my integral is wrong again, since we revert to disk method?
amistre64
  • amistre64
correct; remember shell method gave us a nightmare integral
amistre64
  • 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
  • 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
  • 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
  • anonymous
my bad. recalculating...
amistre64
  • amistre64
((sqrt(6)^5) / 5) - ((12 * (sqrt(6)^3)) / 3) + (36 * sqrt(6)) = 47.0302031 according to the google
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
I put that into my calculator and got -940898/7275....
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=int%28pi%28y^2-6%29^2%29+dy+from+0+to+sqrt%286%29
amistre64
  • amistre64
i mighta typoed it in google lol
anonymous
  • anonymous
no, you have yours right I think. What the fractional form?
anonymous
  • anonymous
it's right!!! thanks a bunch!
anonymous
  • anonymous
I have another problem just like that so I will use this method
amistre64
  • amistre64
:)

Looking for something else?

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