Im having trouble finding the limits of integration i plugged x=6/y into 2x + y = 8 which got me y = -2(6/y) + 8 and then i couldnt finish it finding the y coordinates.
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Intersection points of y = 8 - 2x and y = 6/x:
8 - 2x = 6/x
x(8 - 2x) = x(6/x)
8x - 2x² = 6
2x² - 8x + 6 = 0
2(x² - 4x + 3) = 0
2(x - 1)(x - 3) = 0
x = 1 and x = 3
The y-coordinates for these x-values are
y = 8 - 2(1) = 6
and
y = 8 - 2(3) = 2.
The integral representing the area of the region would be
∫₂⁶ ((4 - y/2) - (6/y)) dy
The integral for the volume of the solid generated by revolving the region about the y-axis (x = 0) is
π ∫₂⁶ ((4 - y/2)² - (6/y)²) dy
8 - 2x = 6/x
x(8 - 2x) = x(6/x)
8x - 2x² = 6
2x² - 8x + 6 = 0
2(x² - 4x + 3) = 0
2(x - 1)(x - 3) = 0
x = 1 and x = 3
The y-coordinates for these x-values are
y = 8 - 2(1) = 6
and
y = 8 - 2(3) = 2.
The integral representing the area of the region would be
∫₂⁶ ((4 - y/2) - (6/y)) dy
The integral for the volume of the solid generated by revolving the region about the y-axis (x = 0) is
π ∫₂⁶ ((4 - y/2)² - (6/y)²) dy
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2x + y = 8
xy = 6
intersections at (1,6) and (3,2)
solve both equations for y:
y = 8 - 2x
y = 6/x
set them equal to find intersections:
6/x = 8 - 2x
3 = 4x - x^2
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
x = 1, x = 3
Volume time:
infinite sum of all the washers rotated around the y axis
integral 2pi (R^2 - r^2) dy
where R = outer radius, r = inner radius
R = 4 - y/2
r = 6/y
integral 2pi((4 - y/2)^2 - (6/y)^2)dy
integral 2pi ((4-y/2)^2-36/y^2) dy
Factor out constants:
= 2pi integral ((4-y/2)^2-36/y^2) dy
Integrate the sum term by term and factor out constants:
= pi integral 2(4-y/2)^2 dy-36 pi integral 1/y^2 dy
For the integrand (4-y/2)^2, substitute u = 4-y/2 and du = -1/2 dy:
= -2 * 2pi integral u^2 du-36 pi integral 1/y^2 dy
The integral of u^2 is u^3/3:
= -2(2 pi u^3)/3-36 pi integral 1/y^2 dy
The integral of 1/y^2 is -1/y:
= (36 pi)/y-(2 pi u^3)/3
Substitute back for u = 4-y/2:
= 2(1/12 pi y (y-8)^3+36 pi)/y
Factor the answer a different way:
= 2(pi (y-2)^2 (y^2-20 y+108))/(12 y)
Which is equivalent for restricted y values to:
= 2pi (y^3/12-2 y^2+16 y+36/y)
evaluate with bounds of intersections [2,6]
2pi ((6)^3/12-2 (6)^2+16 (6)+36/(6)) - 2pi ((2)^3/12-2 (2)^2+16 (2)+36/(2))
which eventually simplifies to 32π/3
yay!
xy = 6
intersections at (1,6) and (3,2)
solve both equations for y:
y = 8 - 2x
y = 6/x
set them equal to find intersections:
6/x = 8 - 2x
3 = 4x - x^2
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
x = 1, x = 3
Volume time:
infinite sum of all the washers rotated around the y axis
integral 2pi (R^2 - r^2) dy
where R = outer radius, r = inner radius
R = 4 - y/2
r = 6/y
integral 2pi((4 - y/2)^2 - (6/y)^2)dy
integral 2pi ((4-y/2)^2-36/y^2) dy
Factor out constants:
= 2pi integral ((4-y/2)^2-36/y^2) dy
Integrate the sum term by term and factor out constants:
= pi integral 2(4-y/2)^2 dy-36 pi integral 1/y^2 dy
For the integrand (4-y/2)^2, substitute u = 4-y/2 and du = -1/2 dy:
= -2 * 2pi integral u^2 du-36 pi integral 1/y^2 dy
The integral of u^2 is u^3/3:
= -2(2 pi u^3)/3-36 pi integral 1/y^2 dy
The integral of 1/y^2 is -1/y:
= (36 pi)/y-(2 pi u^3)/3
Substitute back for u = 4-y/2:
= 2(1/12 pi y (y-8)^3+36 pi)/y
Factor the answer a different way:
= 2(pi (y-2)^2 (y^2-20 y+108))/(12 y)
Which is equivalent for restricted y values to:
= 2pi (y^3/12-2 y^2+16 y+36/y)
evaluate with bounds of intersections [2,6]
2pi ((6)^3/12-2 (6)^2+16 (6)+36/(6)) - 2pi ((2)^3/12-2 (2)^2+16 (2)+36/(2))
which eventually simplifies to 32π/3
yay!