volume of revolution; cyl. shells; CAS; arc length

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soleilion

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1. Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis: y = 1/(4+x^2)^(1/2) , x = -2, x=2, y=0

2. use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis: y = 1/(1+x^2), x = 0, x = 1, y = 0

3. use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the x-axis: x = y^2, y = 1, x = 0

4. use a CAS to find the volume of the solid generated when the region enclosed by y = e^x and y = 0 for x = [1,2] is revolved about the y-axis

5. find the exact arc length of the curve over the stated interval: 24xy = y^4+48 from y = 2 to y = 4

6. find the exact arc length of the parametric curve without eliminating the parameter: x = cos2t, y = sin2t, t = [0, ?/2]

7. find the arc length of the curve between x = -1 and x = 8: curve is y = x^(2/3)

some problems are same, but I dont know how to solve it
thank you!!!
 
Re: Help!!!

soleilion said:
1. Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis
y=1/(4+x^2)^(1/2) , x= -2, x=2, y=0
2. use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis
y=1/(1+x^2), x=0, x=1, y=0

3. use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the x-axis
x=y^2, y=1, x=0

4. use a CAS to find the volume of the solid generated when the region enclosed by y=e^x and y=0 for x=[1,2] is revolved about the y-axis
5. find the exact arc length of the curve over the stated interval
24xy=y^4+48 from y=2 to y=4
6. find the exact arc length of the parametric curve without eliminating the parameter
x=cos2t, y=sin2t, t=[0, ?/2]
7. find the arc length of the curve between x= -1 and x=8
curve is y=x^(2/3)


some problems are same, but I dont know how to solve it
thank you!!!
Click to expand...

Duplicate post - without work:

http://answers.yahoo.com/question/index ... 109AA7awRY

Looks like problems for take-home test
 
soleilion said:
I dont know how to solve it
Click to expand...
If you don't know even how to get started on any of this material, then it would appear that you need to spend a few weeks in diligent study with a qualified tutor. Naturally, this is not a service which can be provided here, and probably "online" is not the way to go. I would strongly recommend that you look for a local tutor. You may have to spend a fair amount (in hourly charges), but I'm afraid that is likely your only choice, other than enrolling in a calculus course at your local college or university.

Note: When you are working with your tutor, you will need to provide him with access to whichever CAS package you are supposed to be using. Since different software works in different ways, the tutor will be able to train you on your particular software only by being able to access it directly.

Good luck! :D

Eliz.
 
I really know how to solve them, but some answers are not match the real answer/. for some answer, I know the formula, but I can not know how to solve the indefinite integral
 
yeah, I understand answers of yahoo
but for this topic, there are 3 more questions
4. I have the function: V=2??xe^x dx from x=[2,1], then I do not know how to fix
6. I have the answer is 2, but real answer is ?
7. the real answer is L= (13?(13)-80?(10)-16)/27, I just do not get 16 part, where is this number come from, I have an answer without 16
 
6. L =??(4[(sin2t)^2+(cos2t)^2])=2??[(sin2t)^2+(cos2t)^2] =2??(sint)^2+(cost)^2 from ?/2 to 0
from second step to third step, I use sin2t=2sintcost, cos2t=(cost)^2-(sint)^2
please tell me where is the error
 
soleilion said:
6. L =??(4[(sin2t)^2+(cos2t)^2])=2??[(sin2t)^2+(cos2t)^2]

=2??(sint)^2+(cost)^2 dt from ?/2 to 0

=2?1. dt from ?/2 to 0


= 2*[?/2 - 0]

= ?

from second step to third step, I use sin2t=2sintcost, cos2t=(cost)^2-(sint)^2
please tell me where is the error
Click to expand...
 
OMG, that mistake is stupid
Im going to be crazy
For #4, I really do not know how to fix
I will check #7again, maybe I made another stupid mistake
thank you
 
7. find the arc length of the curve between x = -1 and x = 8: curve is \(\displaystyle y = x^{\frac{2}{3}}\)
Click to expand...

I assume you know the arc length formula. After differentiating, taking the square root and doing what we need to do:

We must integrate: \(\displaystyle \frac{1}{3}\int_{-1}^{8}\frac{\sqrt{9x^{\frac{2}{3}}+4}}{x^{\frac{1}{3}}}dx\)

It isn't as bad as it looks.

Let \(\displaystyle u=9x^{\frac{2}{3}}+4, \;\ \frac{1}{6}du=\frac{1}{x^{\frac{1}{3}}}\)

Make the subs and finish up.
 
For #7
the question said : explain why formula 4 can not be used to find the arc length of the curve
frmula 4 is ??(1+(dy/dx)^2) dx from a to b

the answer is dy/dx does not exist at x=0

I do not understand
 
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