Find Volume of Revolution by Integrating 1/sqroot(3x+2) around x=0 and x=2

[DeanBH]

so i have y = 1/sqroot(3x+2)

find volume when rotated around x, regions are x=2 and x=0



equation needed: V= integral Pi*y^2*dx

so.

i do intergral pi* (1/sqroot(3x+2))^2 * dx

so i get pi integral 1/(3x+2) dx

so how do i integrate 1/sqroot(3x+2) ?

can someone take me though it , because i know there's meant to be a constant near the ln, but i don't know how to find it.


thanks, sorry if it's explained bad

 

[rock.freak667]

Try putting u=3x+2

 

[nlightner]

To integrate 1/sqroot(3x+2), you can use the substitution method. Let u = 3x+2, then du/dx = 3 and dx = du/3. Substituting this into the integral, we get:

V = pi integral 1/u^(1/2) * (1/3) du

= (pi/3) integral u^(-1/2) du

= (pi/3) * 2u^(1/2) + C

= (2pi/3) * (3x+2)^(1/2) + C

Now, since we are rotating around the x-axis, we need to find the volume between x=0 and x=2. So, we plug in these values into the equation:

V = (2pi/3) * (3(2)+2)^(1/2) - (2pi/3) * (3(0)+2)^(1/2)

= (2pi/3) * (8)^(1/2) - (2pi/3) * (2)^(1/2)

= (2pi/3) * 2^(1/2) - (2pi/3) * (2)^(1/2)

= (2pi/3) * (2^(1/2) - 1)

= (2pi/3) * (2 - 1)

= (2pi/3) * 1

= 2pi/3

Therefore, the volume of revolution is 2pi/3 cubic units.