help with these problems Find area between the two curves y1 = (x - 2)3, y2 = x

Question

help with these problems

Find area between the two curves y1 = (x - 2)3, y2 = x - 2, y1 = x3 - x, y2 = 3x y1 = x(x2 - 3x + 3), y2 = x2, y1 = x2 - 4x + 3, y2 = 3 + 4x - x2 y1 = cos x, y2 = 2 - cos x, 0 x 2pi, y1 = sin(x), y2 = cos(2x),-pi/pi x pi/6 Using appropriate method (disk, washer, shell) to find the volume of the solids generated by revolving the region bounded by the graph of equations about the given lines: (a) the x-axis, (b) the y-axis, c) the line y=8, d) the line x= 6 y = -x + 1, x = 0, x = 1; y1 = x2, y2 = x3 y = sin(x), y = 0, x = 0, x = pi, y1 = x2 - 4x + 3, y2 = 3 + 4x - x2 y1 = 1 + cos x, y2 = 3 - cos x, 0 x 2pi, y = root 25 - x2, x = 0, y = 0. y = 1/x, x = 0, x = 4, y = 0, y = root x + 2, x = 0, y = x. y = 1/2 x2 + 1, x = 0, x = 2, y = 0, (y - 2)2 = 4 - x, x = 0.

Explanation / Answer

I can suggest to you the basic method for solving but solving these many questions would take too too much time

1) i)draw the curves

ii)find their points of intersection

iii) integrate l y1-y2 l i.e. if y1>y2 then y1-y2 else y2-y1 b/w the respective limits

(you might need to break the limits into various parts depending on sign of y1-y2)

iv)integrate and put limits for answer

2) plot the graph and find the limits by visualization of the region

{if you can't then i can tell you some specific one; reply as comment}

a) x- axis : integrate (pi*y^2 dx)   

b) y- axis : integrate (pi*x^2 dy)

c)y=8 : integrate (pi*(y-8)^2dx)

d)x=6 : integrate (pi*(x-6)^2dy)

The integration shouldn't be a problem

You might need to reduce equation solely in form of x i.e. represent y in form of x solely in equations involving dx and vice versa ; square/take square root etc

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