integral (volume)(急!)

The points of intersection between the curves are (-4, 0) and (0, 2).

Also the line x - 2y + 4 = 0 is farther away from the line x = -4 than y2 = x + 4 in the range from y = 0 to y = 2. So the required volume is:

V = π∫(y = 0 → 2) (2y)2 dy - π∫(y = 0 → 2) (y2)2 dy

= π∫(y = 0 → 2) (4y2 - y4) dy

= π[4y3/3 - y5/5] (y = 0 → 2)

= 64π/15 cubic units

approach: consider a strip with thickness dy rotated about x=a, the radius of that strip is (x-a) => volume of strip = (pi) (x-a)^2 (dy)

as the volume is rotated about a vertical line, x = -4
i.e. larger volume about x = -4 minus smaller volume about x = -4
the intersection points are (-4,0) and (0,2)

the volume bounded by x-2y+4 = 0 about the line x = -4,
(pi) Int0>2 [(2y-4) - (-4)]^2 dy
(pi) Int0>2 [4y^2] dy
(4pi/3) [y^3] 0>2
(4pi/3) [0 - 8]
32pi/3 (always takes absolute value)

the volume bounded by y^2 = x+4 about the line x = -4,
(pi) Int0>2 [(y^2 - 4) - (-4)]^2 dy
(pi) Int0>2 [y^4] dy
(pi/5) [y^5] 0>2
(pi/5) [0 - 32]
32pi/5 (always takes absolute value)

the net volume between is [32pi/3] - [32pi/5] = 64pi/15

By disk method or cylindrical shell method.