Volumes of solids using calculus

[darkmaster25]

Regarding a question in MQ 12 Ex 6G q3 if u have the textbook.

Anywhere the question asks:
Calculate the volume generated of the the region bounded by the curve sqrt(x-1), the y axis and the lines y=0 and y=2. I know how to find the area in terms using dy. I just want to know why my way doesn't work. I don't know how to use latex so I attached the screenshot of the equation.

Thanx

[darkmaster25]

Also the answer for this is 206π/15 cubic units, which I got from the first method but not the other method that i posted. The other method got me 12π.

[Bhootnike]

Regarding a question in MQ 12 Ex 6G q3 if u have the textbook.

Anywhere the question asks:
Calculate the volume generated of the the region bounded by the curve sqrt(x-1), the y axis and the lines y=0. I know how to find the area in terms using dy. I just want to know why my way doesn't work. I don't know how to use latex so I attached the screenshot of the equation.

Thanx

is the question not: 3a: Sketch the region bounded by the curve , the y-axis and the lines y = 0 and y = 2.

?

[darkmaster25]

Sorry i meant for (b) and yes including y=2

[b^3]

The reason it won't work is that if you are integrating with respect to x (so dx) then you are rotating around the x-axis, where as if you are integrating with respect to y, (so dy) then you are rotating around the y-axis.

They are two different answers because they are representing to different shapes/volumes.

Also do we have all the question? As it doesn't say which axis to rotate it around?


Also Please refrain from posting urgent in the topic of the thread: Posting Rules

• Avoid putting "Read this need help urgently" or similar in your thread title. Everyone is entitled to the same amount of attention.

Also regarding LaTeX this may help: Re: Spesh Questions

[Bhootnike]

i dont understand where you are coming from with your equation.
-i.e the terminals of 0 , 5, and 1,5.
and why you are find the vol. between two points?


the question was:
3 a) Sketch the region bounded by the curve y = x −1, the y-axis and the lines y = 0
and y = 2.
b) Calculate the volume generated when this region is rotated about the y-axis.

[darkmaster25]

the question said that it was rotated around the y axis. I know we can do in terms of dy, but i wanted to know why we can't do it in terms of dx the way i have suggested. Shouldn't it yield the same result as in both instances  the areas are rotated 360 around an axis?

Also I am using 0 to 5 as at 5 is where the two graphs intersect. So i am finding the volume of the cylinder generated from the rectangle and taking away the volume of the half parabola from 1 to 5.

[paulsterio]

No, the best way to think about this is a parabola, y = x^2.

Rotating it around the y-axis you will get a bowl shape

Rotating around the x-axis you will get a curved-cone sort of shape

[b^3]

No, the best way to think about this is a parabola, y = x^2.

Rotating it around the y-axis you will get a bowl shape

Rotating around the x-axis you will get a curved-cone sort of shape

Like so:

Source: http://mathplotter.lawrenceville.org/mathplotter/MSP/resources/solids/solids

The one on the left is rotated around the y-axis, with the one on the right being rotated around the x-axis (note this is for Paul's example, was already working on the same example when he posted it ).
They are two different shapes, and have two different volumes (as said above).

[darkmaster25]

Okay I understand a bit better, thanks for the help,  . but I still am a little bit confused as to why the area which is only in the first quadrant and surrounded by both the x and y axis spun 360 degrees around the x or the y axis would not have the same area. I mean doesn't the area look like the one in this attachment and spun around either x or y axis shouldn't the area be the same. I know I am wrong but i just want to clear up why. 

Thx

[Special At Specialist]

No, the best way to think about this is a parabola, y = x^2.

Rotating it around the y-axis you will get a bowl shape

Rotating around the x-axis you will get a curved-cone sort of shape

Like so:
(Image removed from quote.)
Source: http://mathplotter.lawrenceville.org/mathplotter/MSP/resources/solids/solids

The one on the left is rotated around the y-axis, with the one on the right being rotated around the x-axis (note this is for Paul's example, was already working on the same example when he posted it ).
They are two different shapes, and have two different volumes (as said above).

Is that really what they look like? I would've thought that rotating around the x-axis would look like the first third of a 3D ellipse and rotating around the y-axis would look like the shape of water when placed in a lemon squeezer.

[darkmaster25]

the one rotated around the x axis should look like two rooves of those circus tents and the one around the y axis looks fine

[b^3]

OK before we go off here, note that the one that is being rotated around the x-axis, is still the region that is enclosed with the y-axis (and the line y=4), the reason I did this is to keep the areas/regions the same so that I could show that the shape when rotated about the different axis is different and will result in a different volume, even though we started with the same region.

It would be different if it were the region of enclosed between the graph, the x-axis, and another line.

I hope that makes sense.

[pi]

the one rotated around the x axis should look like two rooves of those circus tents and the one around the y axis looks fine

They're both fine to me :S

[paulsterio]

If you're only taking the first and 4th quadrants then yes, it will produce the top of a circus tent, but b^3 has taken all 4 quadrants, so his would be like from x = -3 to x = 3, what you're saying is from x = 0 to x = 3