2 Questions (1 Viewer)

[hit patel]

1) Y= 1/ sqrt (x), explain why 18<1/sqrt(1) + 1/ sqrt(2) + 1/ sqrt(3).... + 1/sqrt (100)<19 . The answer I get is different. Please show me how to do this. Please show proper working.

2) Find volume of curve y= Sqrt (1+x) when rotated about y axis ( for 0≤ x ≤ 8) . My answer is 152PI units^3.


Thanks

 

[HeroicPandas]

made a mistke

 

[cineti970128]

I think I figured out the answer for the first one
Btw got 1864/15 pi cubic unit

 

[cineti970128]

1. Draw y=1/sqrt (x) on a number plane
2. Put integer numbers 1, 2, 3, ... 100
3. Draw rectangles on each interval with 1 unit space, and each rectangles should be above the curve , ( bad explanation...)
4. Note that sum of all rectangles equals to 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(100)
5. This area is greater than the area below the curve x=1 to 100
6.Integrated value equals to 18

7. If you draw 1/sqrt(x-1) on the same number plane- you should have figured out something by this stage!!
Now it gets tricky

8.note that the're is an symptote
9. Note that the area below the curve is greater than the sum.
10. Hence the sum is less than
1(to eliminate the problem with the asymptote) + integral (1/sqrt (x-1))dx (x=2!! to 100) = 2sqrt(99) + 1 < 19

The first one solved!!

 

[hit patel]

I think I figured out the answer for the first one
Btw got 1864/15 pi cubic unit

How did you get this? i found the volume of cylinder and then subtracted the volume the curve makes with x axis to find the volume that it makes with y axis.

Ty

1. Draw y=1/sqrt (x) on a number plane
2. Put integer numbers 1, 2, 3, ... 100
3. Draw rectangles on each interval with 1 unit space, and each rectangles should be above the curve , ( bad explanation...)
4. Note that sum of all rectangles equals to 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(100)
5. This area is greater than the area below the curve x=1 to 100
6.Integrated value equals to 18

7. If you draw 1/sqrt(x-1) on the same number plane- you should have figured out something by this stage!!
Now it gets tricky

8.note that the're is an symptote
9. Note that the area below the curve is greater than the sum.
10. Hence the sum is less than
1(to eliminate the problem with the asymptote) + integral (1/sqrt (x-1))dx (x=2!! to 100) = 2sqrt(99) + 1 < 19

The first one solved!!

Ty

 

[braintic]

How did you get this? i found the volume of cylinder and then subtracted the volume the curve makes with x axis to find the volume that it makes with y axis.

I did the same as you and got a different answer again: 599pi/15

Just checking our integrals:
pi times 3^2 times 8 - pi times (int from 1 to 3) (y^4 - 2y^2 +1) dy

 

[hit patel]

I did the same as you and got a different answer again: 599pi/15

Just checking our integrals:
pi times 3^2 times 8 - pi times (int from 1 to 3) (y^4 - 2y^2 +1) dy

But then arent we rotating about the y axis? why did u minus the volume when rotated around the y axis.

 

[braintic]

But then arent we rotating about the y axis? why did u minus the volume when rotated around the y axis.

For starters, the question is VERY poorly worded. Where did you get it from? There is no mention of which region is rotated. I assumed it was the region bounded by the curve and the x-axis, but that was just a guess. All volume questions should specify the region that is rotated, and not just the curve. And what on earth does it mean by 'Find the volume of the curve ...' ? A solid has volume, a curve has only length.

Have you been given the 'correct' answer?

 

[hit patel]

I am sorry but I had included it in there: the rotation around which axis. But i havent been given the correct answer. It says the region rotated around. Please check the post.

 

[cineti970128]

For starters, the question is VERY poorly worded. Where did you get it from? There is no mention of which region is rotated. I assumed it was the region bounded by the curve and the x-axis, but that was just a guess. All volume questions should specify the region that is rotated, and not just the curve. And what on earth does it mean by 'Find the volume of the curve ...' ? A solid has volume, a curve has only length.

Have you been given the 'correct' answer?

i agree I assumed that the area bounded by the curve and the x axis and x=0 and8 is rotated around y axis

 

[braintic]

I am sorry but I had included it in there: the rotation around which axis. But i havent been given the correct answer. It says the region rotated around. Please check the post.

I'm somewhat confused by your post. Your original post says to rotate about the y-axis, and that is what I've done. The axis of rotation wasn't my issue.

The issue is WHICH REGION is being rotated about the y-axis. Probably because you specified 0≤ x ≤ 8, I assumed it was the region bounded by the curve and the x-axis. But the domain itself is not enough to define the region - it is only the domain for the curve itself, and has not been used to define a region.

I just looked at my collection of HSC questions on Integration back to 1995. Almost every volume question specified a region to be rotated. There were three exceptions. In two of those, they were forming a glass or a bowl, so there was only one way of interpreting the region to get that shape. In the other question, it was impossible to misinterpret.

So I need to ask again - which region are we rotating about the y-axis. Is it the region bounded by the curve and the y-axis, or the region bounded by the curve and the x-axis?

 

[hit patel]

The region bounded by the curve y=sqrt(x+1), the x-axis, the y-axis, 0 ≤x ≤8 , is rotated about the y-axis to
form a solid. Find the volume of the solid.

 

[hit patel]

hey
thanks for the help brantic and cineti and all others. I did also get the same answer the second time however I was not sure if I was right because the wording of the question confused me because It didnt make clear which region was to be rotated since it says x axis and y axis both in the same question without making it clear.

Thanks again

 

[Halvat]

The area bounded by the curve y=sqrt(1+x), y axis, x axis and the line x=8, is to be rotated about the y axis.

Generic volume equation: pi*int (x^2) dy

:. pi*int (between 3 and 1) (y^4 - 2y^2 +1) dy

which gives (496*pi)/15

I believe this is correct

 

[braintic]

The area bounded by the curve y=sqrt(1+x), y axis, x axis and the line x=8, is to be rotated about the y axis.

Generic volume equation: pi*int (x^2) dy

:. pi*int (between 3 and 1) (y^4 - 2y^2 +1) dy

which gives (496*pi)/15

I believe this is correct

I'm afraid that is not correct.
You are rotating the region bounded by the curve, the y-axis, and the line y=3.
You are meant to be rotating the region bounded by the curve, the x-axis, the y-axis, and the line x=8.