integral help?

Question

1)an area is covered by the curve y=(x)^2 , y = 4 , x = 3 and x axes. calculate the volume when it is rotate 360 degree through
a)the x axes
b)the y axes
c)x = 3

2)an area is covered by the curve y=(x+3)^2 , y = 4 , x = 0 and x axes. calculate the volume when it is rotate 360 degree through
a)x=0
b)y=0
c)x=-3

3)an area is covered by the curve y=(x-3)^2 , y = 4 , x = 0 and x axes. calculate the volume when it is rotate 360 degree through
a)x=0
b)y=0
c)x=3

this is not my homework...
i read this question from a books a few days ago.. and i got 2 different answers for the questions.. it supposed to be one answer for all of the question when they are rotate at a vertical axes
(x=0 , x=3 , x=-3)

the answer for
1a=2b=3b
1b=2c=3c
1c=2a=3a
but a get different answers

Answer 1

I agree with G.
Make sure that you are familiar with polar coordinates and the different methods of solving this. One such method is to do the integral for cylinderical shells over the regions.
GlidingSquirrel.

Answer 2

you are able to not combine it indefinitely to get an user-friendly purposes, yet you are able to combine it between -infinity and infinity: because of the fact the functionality is even, evaluate int{from 0 to infinity}[(sin(x))/x]dx. combine by ability of aspects, utilising u = a million/x, dv = sin(x)dx to get int{from 0 to infinity}[(sin(x))/x]dx = int{from 0 to infinity}[[a million - cos(x)]/[x^2]]dx. enable a million/x^2 = int{from 0 to infinity}[te^(-tx)dt]. placed that interior the imperative to make a double imperative. Now swap the order of integration interior the double imperative (justified for nonnegative purposes), and compute the imperative.

Answer 3

1a) V = integral(0,2,pi*x^4dx) + integral(2,3,pi*4^2dx)
= [pi*(x^5)/5]{0,2} + [pi*16*x]{2,3}
= (pi * 32/5) + (pi * 16)
= pi * ( 22.4)

1b) V = integral(0,4,pi * 3^2 dy) - integral(0,4,pi * ydy)
= [pi * 9 * y]{0,4} - [pi * (y^2)/2]{0,4}
= (pi * 9 * 4) - (pi * 16/2)
= pi * (28)

1c) V = integral(0,4,pi * (3- sqrt(y))^2dy)
= integral(0,4,pi * (9 - 6sqrt(y) + y)dy)
= pi * [9y - 6*(2/3)*y^(3/2) + (y^2)/2]{0,4}
= pi * (9*4 - 4*8 + 16/2)
= pi * (12)

That'll do for now.

Answer 4

Do your own homework