# Help with volume of revolution questionsWatch

Thread starter 12 years ago
#1
Hi I am stuck with two questions:

1) The area enclosed by the curve y = x^3 + 1, the x-axis and the line x=1.

I know volume = Pi X the intergration of (y2) and filling in the points X=1 and X=0 but i'm not getting the correct answer.....

2) The area enclosed by the curve y = 4x - x^2 and the x-axis. But where is the second x value? When I intergrate do I not have to fill in two values of X?

Thanks very much
0
quote
reply
Thread starter 12 years ago
#2
Hi thanks for helping, for question 1 I got the same answer as you but the answer is 16Pi/7....
0
quote
reply
12 years ago
#3
I believe the answer is correct. It asked for the volume of the area enclosed by the curve and the x-axis. So you have to integrate over the interval [-1, 1], and not [0, 1].
0
quote
reply
12 years ago
#4
(Original post by e-unit)
The answer is wrong.
Ha ha.
y = x^3 + 1 => y^2 = (x^3+1)^2 = x^6+2x^3+1

VOLUME = pi(x^7/7 + x^4/2 + x) between x = -1 and x = 1
VOLUME = pi(1/7+1/2+1) - pi(-1/7+1/2-1)
VOLUME = 16pi/7
0
quote
reply
12 years ago
#5
(Original post by swifty)

1) The area enclosed by the curve y = x^3 + 1, the x-axis and the line x=1.
Ha ha.
y = x^3 + 1 => y^2 = (x^3+1)^2 = x^6+2x^3+1

VOLUME = pi(x^7/7 + x^4/2 + x) between x = -1 and x = 1
VOLUME = pi(1/7+1/2+1) - pi(-1/7+1/2-1)
VOLUME = 16pi/7
Le sigh.

I would neg rep you back, but I usually quite like you - you're normally helpful and intelligent.
0
quote
reply
12 years ago
#6
(Original post by e-unit)
Le sigh.

I would neg rep you back, but I usually quite like you - you're normally helpful and intelligent.
Yh sorry dude about the neg rep. I just believe that you can be passionate about a subject without achieving the top grades. My mechanics teacher who knows M1-M6 and C1-FP3 like the back of his hand got a B at a-level. Does that mean he hates maths or isn't passionate enough? I think the fact he got a 1st in his maths degree and carried on a phd in maths is enough evidence that he's passionate!
0
quote
reply
Thread starter 12 years ago
#7
Hi Widowmaker could you explain to me why you used the value X=-1 and X=1? Even though it said the x-axis and the line x=1?

Thank you!
0
quote
reply
12 years ago
#8
(Original post by swifty)
Hi Widowmaker could you explain to me why you used the value X=-1 and X=1? Even though it said the x-axis and the line x=1?

Thank you!
It's because the curve crosses the x axis at -1. Set the equation equal to 0 and then you can see where the curve crosses the x axis; hence this is the area bounded by the curve and the x-axis. If that makes sense!
0
quote
reply
X

Write a reply...
Reply
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

• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18
• University of East Anglia
UEA Mini Open Day Undergraduate
Fri, 4 Jan '19
• Bournemouth University
Undergraduate Mini Open Day Undergraduate
Wed, 9 Jan '19

### Poll

Join the discussion

Yes (162)
26.87%
No (441)
73.13%

View All
Latest
My Feed

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

Why not re-start the conversation?

### 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.