The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Volumes of Revolution

I have this question which is rather interesting and quite hard IMO.

I have to find the volumes of revolution generated by the regions A and B when they are each rotated about: the x axis and the y axis

If the equation was y=xny=x^n from a to b,then it means I need to find the volume of revolution for:

•

x = a to b, rotated around the x axis, which would be: ∫abπy2 dx\displaystyle\int^b_a \pi y^2 \, dx ⇒\Rightarrow

Unparseable latex formula:

\displaystyle\int^b_a \pi x^2^n \, dx


•

y = a to b, roated around the y axis, which would be: ∫abπx2 dx\displaystyle\int^b_a \pi x^2 \, dx ⇒\Rightarrow ∫abπy2n dx\displaystyle\int^b_a \pi y^\frac {2}{n}\, dx



are those right?

I also have to find this though and this is what I need help with:

•

x = a to b, rotated around the y axis

•

y = a to b, roated around the x axis



How would you find those?
Reply 1
Avatar for TomLeigh
TomLeigh
2
For the last two, if you have the y co-ordinates and need x co-ordinates, just plug the y co-ordinates into the equation to find the relevant x's. (or vice versa)
Reply 2
Avatar for G O D I V A
G O D I V A
OP
11
TomLeigh
For the last two, if you need y co-ordinates and need x co-ordinates, just plug the y co-ordinates into the equation to find the relevant x's.


I have the coordinates, I have to rotate it around the opposite axis. For example, if the function was y=x2y=x^2, most questions would ask for the volume of revolution from x=1 x = 1 to x=3 x = 3 , rotated around the x axis.

My question is asking the exact same thing, except rotated around the y axis
Reply 3
Avatar for TomLeigh
TomLeigh
2
Yes so find out what y is at x=1 and x=3 (y=1 and y=9) and use those y co-ordinates to rotate around y.
Reply 4
Avatar for G O D I V A
G O D I V A
OP
11
I can see the sense in that, and I don't want to undermine your math skills, but are you sure that is right?

Because my book doesn't have answers so by doing this I wouldn't know if I get the right answer
Reply 5
Avatar for TomLeigh
TomLeigh
2
Ah well I could easily be wrong don't worry. I've only just done this chapter but I just used my common sense mixed with the rules we learned in class. If you think about 2 points on an axis say, 3 and 2, with th graph say y=x^2, then you're not going to use the points (3,0) (2,0) as these would make an area of 0, and the only co-ordinates that you can use with these are the x,y co-ordinates, being (3,9) (2,4) If you get me :biggrin:
Reply 6
Avatar for G O D I V A
G O D I V A
OP
11
Anyone else?
Reply 7
Avatar for JoMo1
JoMo1
12
you have a formula for y in terms of x. Presuming you've done integration by substitution you should be able to make this work by making a substitution for y or x depending on the situation. There may well be a quicker way that someone knows of but I believe this should lead you to the answer.
Reply 8
Avatar for G O D I V A
G O D I V A
OP
11
JoMo1
you have a formula for y in terms of x. Presuming you've done integration by substitution you should be able to make this work by making a substitution for y or x depending on the situation. There may well be a quicker way that someone knows of but I believe this should lead you to the answer.



I dont understand
Reply 9
Avatar for Rainfaery
Rainfaery
12
For rotating around the y-axis, it would be π∫abπy2n dx\pi\displaystyle\int^b_a \pi y^\frac {2}{n}\, dx

not

π∫abπy1n2 dx\pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx.
Reply 10
Avatar for G O D I V A
G O D I V A
OP
11
Rainfaery
For rotating around the y-axis, it would be π∫abπy2n dx\pi\displaystyle\int^b_a \pi y^\frac {2}{n}\, dx

not

π∫abπy1n2 dx\pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx.


thats what i thought but if y=x2y=x^2 then y12=xy^\frac {1}{2}=x

π∫aby2 dx\pi\displaystyle\int^b_a y^2\, dx =
Unparseable latex formula:

\pi\displaystyle\int^b_a y^\frac {1}{2}^2\, dx

= y14y^\frac {1}{4}

hence,

π∫abπy1n2 dx\pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx
G O D I V A
thats what i thought but if y=x2y=x^2 then y12=xy^\frac {1}{2}=x

π∫aby2 dx\pi\displaystyle\int^b_a y^2\, dx =
Unparseable latex formula:

\pi\displaystyle\int^b_a y^\frac {1}{2}^2\, dx

= y14y^\frac {1}{4}

hence,

π∫abπy1n2 dx\pi\displaystyle\int^b_a \pi y^\frac {1}{n^2}\, dx



No, when you have a power to a power, they are multiplied together. y12=xy^\frac {1}{2}=x If you square that, you just get y, because (1/2)^2 is just (2/2), which is one. This is just a basic property of indices.
Reply 12
Avatar for G O D I V A
G O D I V A
OP
11
Oh thanks for that.

I still have to find though,

* x = a to b, rotated around the y axis
* y = a to b, roated around the x axis

Related discussions

Latest

Trending

Trending

Articles for you