I am thinking it would be a Hemisphere with radius 2.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 questionTheAnuja55 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
But my question is...TheAnuja55 wrote:Sorry, I don't have the right answer for this, this is y I posted, to get to know.
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.jkaustubh wrote:But my question is...TheAnuja55 wrote:Sorry, I don't have the right answer for this, this is y I posted, to get to know.
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.
Hi Brent,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 KJ,jkaustubh wrote:Hi Brent,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
Can you put some light on the question, I mean what should one do if one gets such a question??
Thanks
KJ
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/5jkaustubh wrote:I think there is some information missing in the questionTheAnuja55 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
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.
If you were asked to plot x= sqrt(2) what would it look like ?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
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.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
Hi,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
