A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
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
 5 years ago
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?

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you can do areas of circles then.... pi {S} [sqrt(x)]^2 dx ; [0,6]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi x^2  at 6 = 18pi 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sooo close... i forgot to read the hwhole thing

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since you are at the line x=6; subtract 6 from the function to bring it down to x = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1over down..someplace; same idea

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sqrt(x+6) is the function you use to move everything to the x = 0 line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay....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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's the graph I have, yep

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1let me redo what I said lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi {S} [sqrt(x+6)]^2 dx ; [0,6] pi {S} x+6 dx ; [0,6] pi(x^2/2 + 6x) at x = 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you added 6 under the sqrt sign. How do you know to do that?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1thats 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
 5 years ago
Best ResponseYou've already chosen the best response.0OKAY. wow, brain lapse.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we dont want to cahng ethe value of the function; just move it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and I steered wrong again i think; I sppun it around the x axis in that..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you used the disk method to solve the problem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you 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
 5 years ago
Best ResponseYou've already chosen the best response.1when I just did it I spun it around the x axis like an idiot lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12pi {S} x(sqrt(x6)) dx is good too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using shell method would leave me with \[2\pi \int\limits_{0}^{6}(6x)(\sqrt(x+6)dx ?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1maybe harder; so lets reinterpret the graph for y instead of x... does that make sense?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1not 6x; just x; the radius is x as x moves from 0 to 6, but that integral looks scary

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sqrt(x^2(x+6)) = sqrt(x^3 + 6x^2) and integrating that looks painful lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{0}^{6}y(y^2+6)dy ?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x = y^26 is a good tranlation dont you think?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, y^2 minus 6. But the rest of the integration looks okay?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1hmm.... if we shell it with respect to y, that doesnt get us out spin around the x=0 part

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes; but we need a new interval; one that works along the y line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now i am confused...can we try again with x? That makes the most sense to me

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when x = 6 to begin with; y = sqrt(6) so our interval is along the y axis from 0 to sqrt(6)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if we do it with respect to x; we run into an awful integral.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, now I understand. The limits were wrong

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if we do it with respect to y; its easier, we just have to know how far to go onthe y to match

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1at x= 6, our original; y = sqrt(6) so our boundary is from 0 to sqrt(6) of y^26

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and it's minus 6 because of "topbottom"?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{0}^{\sqrt(6)}y(y^26)dy\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1now we do the disc method of areas of circles..... to spin around x=0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pi {S} [y^26]^2 dy; from [0,sqrt(6)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, so my integral is wrong again, since we revert to disk method?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1correct; remember shell method gave us a nightmare integral

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we 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
 5 years ago
Best ResponseYou've already chosen the best response.1pi {S} (y^4 12y^2 +36) dy ints to: pi(y^5/5 12y^3/3 + 36y) at y = sqrt(6) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah; 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
 5 years ago
Best ResponseYou've already chosen the best response.0my bad. recalculating...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1((sqrt(6)^5) / 5)  ((12 * (sqrt(6)^3)) / 3) + (36 * sqrt(6)) = 47.0302031 according to the google

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I put that into my calculator and got 940898/7275....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=int%28pi%28y^26%29^2%29+dy+from+0+to+sqrt%286%29

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i mighta typoed it in google lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you have yours right I think. What the fractional form?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's right!!! thanks a bunch!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have another problem just like that so I will use this method
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.