Integration help: Finding the volume. (1 Viewer)

[old.skool.kid]

Hey guys i need help with these two questions:

1.The area bounded by the curve y=x^2, y=(x+2)^2 and the x-axis is rotated about the x-axis. Find the volume of the solid of revolution formed.

2.The area enclosed btween the curves y=x^2 and y=x^3 is rotated about the x-axis. Find the volume of the solid formed.

I was thinking you had to sub them in to each other to find their poi but i couldnt even do that for 1. For 2. i found the poi but then i wasnt sure what to do from there.

Thanks for your help.

 

[Timothy.Siu]

umm
i'm not too sure about the first one but i did
integration of y^2 which in this case is (x²-(x+2)²)²=16x²+32x+16
then i integrated that with the limits 0 and -2 (thats where it intersects with the x-axis) with pi outside. and i ended up with 10 2/3 pi but i did it again using another way and i got 62/5 pi

for the second one y²= (x²-x³)
then i integrated that and i got 1/105pi

 

[shaon0]

Hey guys i need help with these two questions:

1.The area bounded by the curve y=x^2, y=(x+2)^2 and the x-axis is rotated about the x-axis. Find the volume of the solid of revolution formed.

2.The area enclosed btween the curves y=x^2 and y=x^3 is rotated about the x-axis. Find the volume of the solid formed.

I was thinking you had to sub them in to each other to find their poi but i couldnt even do that for 1. For 2. i found the poi but then i wasnt sure what to do from there.

Thanks for your help.

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 [0.2x^5+2x^4+8x^3+16x^2+16x)-(0.2x^5)][limits:0 to -1]
pi [2x^4+8x^3+16x^2+16x][limits:0 to -1]
pi (2/5)
2pi/5 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^6 dx - S [limits: 1,0] x^4 dx)
abs(pi [x^7/7][limits: 1,0] - [x^5/5][limits: 1,0])
abs(pi [1/7 - 1/5])
abs(-2pi/35)
2pi/35 units^3

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

 

[tommykins]

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

Watch that again :]

 

[shaon0]

Watch that again :]

did i make a mistake? Damn, i should review volumes again. sorry

 

[lyounamu]

1) 2/5 pi

You should separate the area and find the volume for each part.

Then you get this.

 

[tommykins]

Yes lyounamu - that's what I got.

The easiest way to do q1) is to observe that the area from 0-> -1 in y = x^2 is the same area from -1 -> -2 in y = (x+2)^2, so basically you just need to integrate y^2 = x^4 from 0->-1 and multiply it by 2.

 

[lyounamu]

Yes lyounamu - that's what I got.

The easiest way to do q1) is to observe that the area from 0-> -1 in y = x^2 is the same area from -1 -> -2 in y = (x+2)^2, so basically you just need to integrate y^2 = x^4 from 0->-1 and multiply it by 2.

And the areas that others got is too high. Logically it should be really smaller than 3 units.

 

[tommykins]

And the areas that others got is too high. Logically it should be really smaller than 3 units.

Yep, I thought they answered it fine until I saw 6pi³ and was like woah, too big.

I did it by slices though

 

[shaon0]

Yep, I thought they answered it fine until I saw 6pi³ and was like woah, too big.

I did it by slices though

i didn't want to use annulus discs or any 4unit techniques so i just did whatever i could remember. plus that technique is too hard to do in your head quickly.

 

[tommykins]

It turns out to be the same however, I did it normally with 3unit techniques and then used 4unit techniques to find it was the same.

 

[shaon0]

It turns out to be the same however, I did it normally with 3unit techniques and then used 4unit techniques to find it was the same.

what mistake did i make?

 

[lyounamu]

what mistake did i make?

You only used limits 0 to -1.

 

[tommykins]

Your limits for (x+2)^2 is wrong.

V = pi [ (int -1 -> -2 (x+2)^4 dx) + (int 0->-1 x^4 dx)]

 

[shaon0]

Your limits for (x+2)^2 is wrong.

V = pi [ (int -1 -> -2 (x+2)^4 dx) + (int 0->-1 x^4 dx)]

how did you get those limits?

 

[lyounamu]

how did you get those limits?

If you draw a graph containing both functions you will see that the area between them are under each graph. Then you use the integeral to work that out each part out.

 

[lyounamu]

umm
i'm not too sure about the first one but i did
integration of y^2 which in this case is (x²-(x+2)²)²=16x²+32x+16
then i integrated that with the limits 0 and -2 (thats where it intersects with the x-axis) with pi outside. and i ended up with 10 2/3 pi but i did it again using another way and i got 62/5 pi

for the second one y²= (x²-x³)
then i integrated that and i got 1/105pi

You have to work that out separately.

2. 2/35 pi

 

[shaon0]

If you draw a graph containing both functions you will see that the area between them are under each graph. Then you use the integeral to work that out each part out.

Okay...i did it in my head so i didn't really visualise the volume.
Thanks for your help

 

[tommykins]

how did you get those limits?

Drawing the graph and by observation that the area from -2 -> -1 is the same as -1 -> 0 for both graphs.

 

[shaon0]

Drawing the graph and by observation that the area from -2 -> -1 is the same as -1 -> 0 for both graphs.

alrite thanks