Calculating the volume of the rotated region?

Question

I need help solving this Calculus problem using integrals, a disk or washer method:

y = x^2-2x ; y = -4x+3
about the line y = -3

The directions say to write the integral needed to calculute the volume generated by rotating the region bounded by the given functions about the given line, then calculate the volume of the given solid. Thanks for any help

Answer 1

First find the domain that the area is found. Set the equations equal to each other.



x^2-2x=(-4)x+3 ------> x^2+2x-3=0 -------> (x+3)(x-1)=0



So a=(-3) and b=1 for the domain of the integral. Now find the equation on top and on bottom by plugging in a value between the domain.



(0)^2-2(0)=0 -4(0)+3=3 Since (-4)x+3 has the higher value, it's the top equation.



To rotate a solid, slightly different steps than normal must be taken. The basic formula is:



(pie)(int from a-b): [top eqn-(line being revolved around)]^2 - [bottom eqn-(line being revolved around)]^2



So:



(pie)(int from -3 to 1): [(-4)x+3-(-3)]^2 - [x^2-2x-(-3)]^2

Answer 2

Okay well the first thing that you need to do is figure out the limits on the integral. The limits are going to be the points of intersection of the two functions. These are x=1 and x=-3 (You get these by setting the two equations equal to each other and solving). Okay so in this case you will be using the washer method which is pi*ʃ(R^2-r^2)dx, where R is the function that would be on the "outside" when the region rotates around and r is the function that would be on the inside. So this gives you pi*ʃ[(x^2-2x)^2-(-4x+3)^2]dx with the limits on the integral being -3 to 1. Sorry it is kinda of hard to write this stuff properly with a keyboard

Answer 3

a million) that's finished with the aid of integration in spite of the undeniable fact that it is not needed. The quickly line in the process the beginning will produce a cone while revolved approximately each and each axis. around x axis the cone has radius 5 and top a million so V = (a million/3)*pi*25*a million = 25pi/3. around y axis the cone has radius a million and top 5 so V = (a million/3)*pi*a million*5 = 5pi/3. 2. This does want integration. around x axis V = pi*INT y^2 dx = pi*INT [0, 4] x dx = pi*(a million/2)(x^2) [0, 4] = 8pi. around y axis V = 2pi*INT xy dx = 2pi*INT [0, 4] x^3/2 dx = 2pi*((2/5)x^5/2 [0, 4] = 128pi/5. P.S. with the aid of 8pi*u^3 do you recommend instruments cubed? that's usually ignored.