Volumes question >.< (1 Viewer)

[xMrRand0m]

Hi guys, I need help with a volumes question. Any help would be appreciated.
Thanks in advance =)

Find the volume when the area bounded by the y=-x^2-3x+6 and x+y-3=0 is rotated about x=3.

 

[alakazimmy]

You sure your equations are correct? The parabola and the line never intersect, so no area can be bounded.

 

[Timothy.Siu]

yeah they do, at x=1 or -3

 

[alcalder]

Actually, yes they do can. The curves cross at (-3,6) and (1,2).

y = - x² - 3x + 6
x + y - 3 = 0

Make cylindrical shells in width dx and height

Larger cylindrical shell height
h_large = (- x² - 3x + 6) - (-x+3)
h_large = - x² - 2x + 3
Smaller cylindrical shell height
h_small = (- (x+dx)² - 3(x+dx) + 6) - (-(x+dx)+3)
h_small = - (x+dx)² - 2(x+dx) + 3

Radius larger shell = y = 3-x
Radius smaller shell = 3-(x+dx)

Thus cylindrical shell volume
V-v then do your limits from x = -3 to x=1

Answer = 47 1/3 pi units³

but I could be wrong because it has been a while since I've done this.

 

[alakazimmy]

oops, read the pluses as minuses.

 

[Timothy.Siu]

lol i probably got it wrong,

i got 85 1/3 pi

 

[99uaifohshiz]

is this part of 4U maths??

sounds interesting

 

[xMrRand0m]

Thanks to all those answers; it turns out the answer's 256/3 pi = 85 1/3 pi and Timothy got that right. Could you please post your working out?

Thankyou =)

edit: 99uaifohshiz; yeah it is 4U maths. lol.

 

[Timothy.Siu]

Thanks to all those answers; it turns out the answer's 256/3 pi = 85 1/3 pi and Timothy got that right. Could you please post your working out?

Thankyou =)

edit: 99uaifohshiz; yeah it is 4U maths. lol.

wow that was lucky
umm
i did it by cylindrical shells,
so if u draw a diagram, the radius of the shell is 3-x and the height of the shell is (-x^2-2x+3)
so the shell is 2pi (3-x)(-x^2-2x+3) dx (something like that)
it intersects at x=1,-3
so,
...../1
2pi| (3-x)(-x^2-2x+3)dx
..../-3

basically u expand and integrate and sub in those limits and u shud get it, lol my explanations aren't very good but hope u get it

 

[xMrRand0m]

wow that was lucky
umm
i did it by cylindrical shells,
so if u draw a diagram, the radius of the shell is 3-x and the height of the shell is (-x^2-2x+3)
so the shell is 2pi (3-x)(-x^2-2x+3) dx (something like that)
it intersects at x=1,-3
so,
...../1
2pi| (3-x)(-x^2-2x+3)dx
..../-3

basically u expand and integrate and sub in those limits and u shud get it, lol my explanations aren't very good but hope u get it

Haha awesome! Thanks. I get it now! =)