Find the volumes of the solids generated by revovling the region in the first quadrant bounded by the curve x=4y-4y^3 and the y-axis about the given axes:
a. the x-axis
b. the line y=1
NO idea how to set it up!!! HElP!!
I did that already and got it incorrect
No problem! I can help you set up the problem.
To find the volume of the solid generated by revolving the region bounded by the curve and the given axes, we'll use the method of cylindrical shells.
a. When revolving the region about the x-axis:
- First, we need to find the limits of integration for y. To find these limits, we set the given equation, x = 4y - 4y^3, equal to 0 and solve for y. This gives us two x-intercepts: y = 0 and y = 1.
- The limits of integration for y are from 0 to 1.
- We need to express x in terms of y: x = 4y - 4y^3.
- The radius of each cylindrical shell is the x-coordinate, which is x = 4y - 4y^3.
- The height of each cylindrical shell is the difference between the x-axis and the curve, which is h = x.
- Now the volume of each cylindrical shell can be calculated as: V = 2πrh = 2π(4y - 4y^3)(4y - 4y^3) = 2π(4y - 4y^3)^2.
- Finally, integrate this expression from 0 to 1: ∫(0 to 1) 2π(4y - 4y^3)^2 dy to find the volume.
b. When revolving the region about the line y = 1:
- Similar to the previous case, we need to find the limits of integration for y, which are again from 0 to 1.
- The radius of each cylindrical shell now becomes the distance between the curve and the line y = 1, which is r = 1 - (4y - 4y^3).
- The height of each cylindrical shell is h = x.
- The volume of each cylindrical shell is V = 2πrh = 2π(1 - (4y - 4y^3))(4y - 4y^3).
- Integrate this expression from 0 to 1: ∫(0 to 1) 2π(1 - (4y - 4y^3))(4y - 4y^3) dy to find the volume.
Remember to evaluate the integrals to find the exact values of the volumes.
Let me know if you need any further assistance with the calculations!
the function is
x = 4y(1-y^2)
so you know it crosses the y-axis at -1,0,1
So you're dealing with the bump in QI from y=0 to 1
To revolve that about the x-axis, I'd suggest shells, so
v = integral[0,1] 2pi r*h dy
where r = y, and h = x
v = 2pi integral[0,1] y(4y-y^3) dy
= 2pi Integral 4y^2 - y^4 dy
= 2pi (4/3 y^3 - 1/5 y^5) [0,1]
= 2pi (4/3 - 1/5)
= 34pi/5
Around y=1, same formula, but now r = 1-y.