region enclosed by the y-axis

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TheAnuja55: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Wed Jul 27, 2011 12:35 amFollowed by: 2 members

by TheAnuja55 » Wed Nov 07, 2012 10:31 pm

12. If the region enclosed by the y-axis, the line y=2 , and the curve y=sqrt(2) is revolved about the y-axis, the volume of the solid generated is

(A) 32PI/5
(B) 16PI/3
(C) 16PI/5
(D) 8PI/3
(E) PI

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Source: Beat The GMAT — Problem Solving | Wed Nov 07, 2012 10:31 pm

FLUID: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Thu Nov 01, 2012 7:52 pm; Thanked: 7 times

by FLUID » Wed Nov 07, 2012 11:39 pm

TheAnuja55 wrote:12. If the region enclosed by the y-axis, the line y=2 , and the curve y=sqrt(2) is revolved about the y-axis, the volume of the solid generated is

(A) 32PI/5
(B) 16PI/3
(C) 16PI/5
(D) 8PI/3
(E) PI

I am thinking it would be a Hemisphere with radius 2.

=> 2/3 * pi * (2)^3 = 16PI/3 (Choice B).

can you comment on right answer

Thanks,

"Take Risks in Your Life.
If you Win, you can Lead!
If you Loose, you can Guide!".

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jkaustubh: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Mon Aug 13, 2012 11:53 pm; Location: Mumbai; Thanked: 1 times

by jkaustubh » Thu Nov 08, 2012 1:28 am

TheAnuja55 wrote:12. If the region enclosed by the y-axis, the line y=2 , and the curve y=sqrt(2) is revolved about the y-axis, the volume of the solid generated is

(A) 32PI/5
(B) 16PI/3
(C) 16PI/5
(D) 8PI/3
(E) PI

I think there is some information missing in the question

Y axis is X=0

Y=2 and Y=sqrt(2) both are straight lines.

these three lines don't form any enclosed region.

Please clarify.

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TheAnuja55: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Wed Jul 27, 2011 12:35 amFollowed by: 2 members

by TheAnuja55 » Thu Nov 08, 2012 1:34 am

Sorry, I don't have the right answer for this, this is y I posted, to get to know.

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jkaustubh: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Mon Aug 13, 2012 11:53 pm; Location: Mumbai; Thanked: 1 times

by jkaustubh » Thu Nov 08, 2012 2:02 am

TheAnuja55 wrote:Sorry, I don't have the right answer for this, this is y I posted, to get to know.

But my question is...

Y=2 and Y=sqrt(2) both are straight lines.
also the Y axis dont have any enclosed boundry.

Then how can three infinite lines with no enclosed boundary, form a three dimensional space with finite volume on rotation along the Y axis.

Hope you understand my question.

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TheAnuja55: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Wed Jul 27, 2011 12:35 amFollowed by: 2 members

by TheAnuja55 » Thu Nov 08, 2012 5:07 am

jkaustubh wrote:
TheAnuja55 wrote:Sorry, I don't have the right answer for this, this is y I posted, to get to know.
But my question is...

Y=2 and Y=sqrt(2) both are straight lines.
also the Y axis don't have any enclosed boundary.

Then how can three infinite lines with no enclosed boundary, form a three dimensional space with finite volume on rotation along the Y axis.

Hope you understand my question.

Yes, there are only two infinite lines, which never gonna enclose the boundaries. So I found that question flawed. But this was given in the list of 700+ questions from gmatclub, that's why I needed opinion about it.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Nov 08, 2012 6:58 am

This looks like a classic integral calculus question (unless I'm missing something).
As such, it is WAYYYYYYYYYYYYYYYYYYYY out of scope for the GMAT.
I suggest that you ignore this question and move on to questions that you may actually see on test day.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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jkaustubh: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Mon Aug 13, 2012 11:53 pm; Location: Mumbai; Thanked: 1 times

by jkaustubh » Thu Nov 08, 2012 10:43 am

Brent@GMATPrepNow wrote:This looks like a classic integral calculus question (unless I'm missing something).
As such, it is WAYYYYYYYYYYYYYYYYYYYY out of scope for the GMAT.
I suggest that you ignore this question and move on to questions that you may actually see on test day.

Cheers,
Brent

Hi Brent,

Can you put some light on the question, I mean what should one do if one gets such a question??

Also, if we consider a small part of width dx...located at a distance x from the Y axis then we need to integrate the following expression from -infinite to +infinite:

pi*h*x*x*dx

on integrating we get

[pi*h*(x^3)/3] with limits as -infinite to +infinite.

how to proceed from here??

Thanks
KJ

Last edited by jkaustubh on Thu Nov 08, 2012 10:50 am, edited 1 time in total.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Nov 08, 2012 10:45 am

jkaustubh wrote:
Brent@GMATPrepNow wrote:This looks like a classic integral calculus question (unless I'm missing something).
As such, it is WAYYYYYYYYYYYYYYYYYYYY out of scope for the GMAT.
I suggest that you ignore this question and move on to questions that you may actually see on test day.

Cheers,
Brent
Hi Brent,

Can you put some light on the question, I mean what should one do if one gets such a question??

Thanks
KJ

Hi KJ,

My point in all of this is that you won't get a question like this on the GMAT. The skills required to solve this question are beyond the scope of the GMAT.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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cwilliams66: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Apr 21, 2016 9:48 am

by cwilliams66 » Thu Apr 21, 2016 9:58 am

jkaustubh wrote:
TheAnuja55 wrote:12. If the region enclosed by the y-axis, the line y=2 , and the curve y=sqrt(2) is revolved about the y-axis, the volume of the solid generated is

(A) 32PI/5
(B) 16PI/3
(C) 16PI/5
(D) 8PI/3
(E) PI
I think there is some information missing in the question

Y axis is X=0

Y=2 and Y=sqrt(2) both are straight lines.

these three lines don't form any enclosed region.

Please clarify.

y=sqrt(2) is not a straight line. There is an enclosed region formed and you have to use the calculus equation for the revolution of shapes in terms of Y (since it is revolved around the y-axis). Volume of the solid = pi x INTEGRAL(from 0 to 2) of (y^2)^2 dy and it will give you answer A. 32pi/5

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Thu Apr 21, 2016 10:19 am

cwilliams66 wrote:

y=sqrt(2) is not a straight line. There is an enclosed region formed and you have to use the calculus equation for the revolution of shapes in terms of Y (since it is revolved around the y-axis). Volume of the solid = pi x INTEGRAL(from 0 to 2) of (y^2)^2 dy and it will give you answer A. 32pi/5

If you were asked to plot x= sqrt(2) what would it look like ?

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Thu Apr 21, 2016 10:23 am

Let me help you out:

https://www4f.wolframalpha.com/Calculat ... ge/gif&s=8

This is a plot of y=sqrt(2) from Wolfram Alpha

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cwilliams66: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Apr 21, 2016 9:48 am

by cwilliams66 » Thu Apr 21, 2016 8:55 pm

regor60 wrote:Let me help you out:

https://www4f.wolframalpha.com/Calculat ... ge/gif&s=8

This is a plot of y=sqrt(2) from Wolfram Alpha

Ahhh yes, my (embarassing) mistake. I was reading the question off of my worksheet which says y= sqrt(x) rather than sqrt(2) as suggested in this post.. Now I understand the confusion. I believe the problem was supposed to read sqrt(x) and not sqrt(2). Thanks for calling out my mistake.

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chetan.sharma: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Sun Dec 27, 2015 7:03 amFollowed by: 1 member; Thanked: 8 times

by chetan.sharma » Fri Apr 22, 2016 7:25 pm

TheAnuja55 wrote:12. If the region enclosed by the y-axis, the line y=2 , and the curve y=sqrt(2) is revolved about the y-axis, the volume of the solid generated is

(A) 32PI/5
(B) 16PI/3
(C) 16PI/5
(D) 8PI/3
(E) PI

Hi,
firstly the Q is incomplete as also mentioned above by kaustabh..
y=square root 2, is not a curve..
I believe it should be y = sqroot2 * x...
so, when you draw the three lines..
one end is y-axis..
second end is y=2
and third is curve y= sqroot2*x..

so answer is 1/4 of a sphere..
ans is 1/3 pi r^3 = 8/3 pi r^3

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