Volume of Region A in First Quadrant Rotated Around y = -2

[anthonym44]

[SOLVED] Volumes of Revolution

Homework Statement

find the volume of region A in the first quadrant that is inclosed in the parabola 2x(2-x) and the x-axis, of which is rotated around the axis y = -2.

Homework Equations

y=(4+2x^2) y= -2 piS(from 0 to 2) (R^2 - r^2)

The Attempt at a Solution

piS(0 to 2)(4x + 2x^2 +2)^2 + (2^2)
piS(0 to 2) (4x^4 + 16x^3 + 24x^2 + 8x + 4) = (4x^3/3 + 16x^4/4 + 24x^3/3 + 8x^2/2 + 4x)|(0 to 2) from this you can plug in 2 to get 158pi obviously this is too large to be the area thus my dilemma. any help is appreciated thanks.

 

[HallsofIvy]

What you've written is extremely hard to read- even after I figured out that "S" indicates the integral.

In any case you can do this by "washers". For each x, between 0 and 2, the upper boundary is y=2x(2-x)= 4x- 2x². That's the distance from a point on the upper boundary to the y-axis. It's distance from y= -2 is 4x- 2x^{2+ 2 and that is the radius of the circle it makes about the line y= -2. The area of the disk it sweeps out is $\pi(2+ 4x- 2x^2)^2$. The "inner disk" is swept out by points on the x-axis which has constant distance 2 from y= -2. It's area is $4\pi$. Taking "dx" to represent the thickness, the volume of each washer is $\pi ((2+ 4x- 2x^2)^2- 4)dx$ and so the integral is
$$\pi \int_0^2 (2+ 4x- 2x^2)^2- 4)dx= \pi\int_0^2 4x^4- 16x^3+ 8x^2+ 16x dx$$. I get 224/15.-}

 

[anthonym44]

thanks a lot, sorry about the difficulty you had while reading this. How do you right out the equations like you have it? I would like to be able to do that so my posts later on are more clear. Thanks again for your help.