# integration questionWatch

#1
The arc of the parabola with equation y²=4ax, where a is a positive constant, from the origin O to A (a, 2a) is rotated through 2π radians about the x-axis. How do I show that the area of the surface generated is:

8πa² (2√2-1) / 3

??
0
quote
12 years ago
#2
y^2 = 4ax => dy/dx = 2a/y = a/sqrt(ax)

So that:
sqrt(1 + (dy/dx)^2) = sqrt(1 + a/x)

Then:
surface area = 2pi ∫ y sqrt(1 + (dy/dx)^2) dx
= 2pi ∫ 2 sqrt(ax) sqrt(1 + a/x) dx
= 4pi ∫ sqrt(ax + a^2) dx
= 4pi a ∫(x + a)^(1/2) dx
= 4pi a [(2/3) (x + a)^(3/2)]{0, a}
= (8/3)pi a [(2a)^(3/2) - a^(3/2)]
= (8/3)pi a^2 (2sqrt(2) - 1)
0
quote
#3
thanks dvs
0
quote
12 years ago
#4
(Original post by dhokes)
The arc of the parabola with equation y²=4ax, where a is a positive constant, from the origin O to A (a, 2a) is rotated through 2π radians about the x-axis. How do I show that the area of the surface generated is:

8πa² (2√2-1) / 3

??
Please post future mathematical problems in the Maths section of Academic Help.
0
quote
12 years ago
#5
(Original post by Angel Interceptor)
Please post future mathematical problems in the Maths section of Academic Help.
0
quote
X

new posts

Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Sheffield Hallam University
Thu, 13 Dec '18
• University of Buckingham
Thu, 13 Dec '18
• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18

### Poll

Join the discussion

Yes (51)
26.56%
No (141)
73.44%