Integration help: Finding the volume. (1 Viewer)

[old.skool.kid]

Hey thanks for all your help guys..
I was able to get 1. using your limits Shaon0. Maybe you did something wrong in here:
"pi [2x^4+8x^3+16x^2+16x+32/5][limits:0 to -1]"

So this is what i did:
V=pi S[0,-1] (x+2)^4 - S[0,-1] x^4
=pi[(x+2)^5/5] - [x^5/5]
=pi 1/5--1/5
=2pi/5

 

[lyounamu]

Hey thanks for all your help guys..
I was able to get 1. using your limits Shano. Maybe you did something wrong in here:
"pi [2x^4+8x^3+16x^2+16x+32/5][limits:0 to -1]"

So this is what i did:
V=pi S[0,-1] (x+2)^4 - S[0,-1] x^4
=pi[(x+2)^5/5] - [x^5/5]
=pi 1/5--1/5
=2pi/5

Yep, that seems right except the limit parts. You will have to use limit (-1, -2) because the area is under that limit. You got the same answer as us because the area is identical from 0 to -1 to -1 to -2. But you will potentailly lose a mark for that part.

 

[tommykins]

Hey thanks for all your help guys..
I was able to get 1. using your limits Shaon0. Maybe you did something wrong in here:
"pi [2x^4+8x^3+16x^2+16x+32/5][limits:0 to -1]"

So this is what i did:
V=pi S[0,-1] (x+2)^4 - S[0,-1] x^4
=pi[(x+2)^5/5] - [x^5/5]
=pi 1/5--1/5
=2pi/5

Your limits are wrong.

 

[old.skool.kid]

1. y=x^2, y=(x+2)^2
Thus, 4x+4=0....x=-1
pi S ([limits:0 to -1] (x+2)^4 dx - S [limits:0 to -1] x^4 dx)
pi [(0.2(x+2)^5) - (0.2x^5)][limits:0 to -1]
pi [2x^4+8x^3+16x^2+16x+32/5][limits:0 to -1]
pi (32/5 - 2/5]
6pi units^3

2. y=x^2 and y=x^3
x^3=x^2
x^3-x^2=0
x^2(x-1)=0
Thus, limits are; 1,0

abs(pi S [limits: 1,0] x^3 dx - S [limits: 1,0] x^2 dx)
abs(pi [0.25x^4][limits: 1,0] - [x^3/3][limits: 1,0])
abs(pi [1/4 - 1/3])
abs(-pi/12)
pi/12 units^3

I think these are the solutions although i haven't done volumes of revolution in ages.

Thats not the right answer. I think you forgot to make it y^2, but i still cant do it. Can anyone do 2?

 

[lyounamu]

Thats not the right answer. I think you forgot to make it y^2, but i still cant do it. Can anyone do 2?

I did 2. Look above.

Here is the working out:

y = x^2 and y= x^3

They intersect at x=1, y=1 and x=0, y=0

So the lims are 0 to 1.

Now, you work out the volume separately.

y^2 = x^4 and y^2 = x^6

I ( x^4) = pi [x^5/5] and I (x^6) = pi [x^7/7] and the lims are 0 to 1.

So V1 = 1/5 pi and V2 = 1/7 pi

V1 - V2 = 2/35 pi

 

[tommykins]

For 2. y = x^2 is on top of y = x^3.


Volume is pi int 1->0 x^4 - x^6 dx = pi [ x^5/5-x^7/7] 1->0 = 2pi/35

 

[shaon0]

For 2. y = x^2 is on top of y = x^3.


Volume is pi int 1->0 x^4 - x^6 dx = pi [ x^5/5-x^7/7] 1->0 = 2pi/35

Yeah i got it.

 

[old.skool.kid]

Ahh yep, i just made a stupid mistake..

Thanks again..