A PIE WITH CIRCULAR CROSS SECTIONS IS 24 CM ACROSS THE TOP , 16 CM ACROSS THE BOTTOM, AND 6 CM DEEP. FIND THE VOLUME OF THIS DISH? Solve using integrals
Can some one please help me with this question? I am very much lost =(
Can some one please help me with this question? I am very much lost =(
-
y = 3/2 x - 12 <== represents outer side of pie
y = 6 <== top of pie
y = 0 <== bottom of pie
x = 0 <== center
Volume = Area bounded by the above four equations rotated about the y axis.
V = π integral (a to b) [ f(y)^2 ] dy
y = 3/2 x - 12
x = 2/3 y + 8
V = π * integral (0 to 6) [ (2/3 y + 8)^2 ] dy <=== integral to evaluate
y = 6 <== top of pie
y = 0 <== bottom of pie
x = 0 <== center
Volume = Area bounded by the above four equations rotated about the y axis.
V = π integral (a to b) [ f(y)^2 ] dy
y = 3/2 x - 12
x = 2/3 y + 8
V = π * integral (0 to 6) [ (2/3 y + 8)^2 ] dy <=== integral to evaluate
-
Insufficient information without a picture. Are the sides of the pie straight segments?
Assuming they are, let's use cylindrical coordinates (p, theta, z), and center the pie on the origin, resting on top of the z=0 plane. We will integrate by disks. (Could go by shells, too, doesn't matter.)
You can set up the triple integral for
p dp d(theta) dz
where the limits are p=0 to radius
theta goes from 0 to 2 pi (all the way around)
z goes from 0 to 6
But what is the radius?
At z=0, p (rho, the radius) is 8. at z=6, p=12. You can derive the formula that p = 8 + 2z/3
so the limits of integration for p are 0 and 8 + 2z/3
Is that enough, can you do the integrals from there?
Assuming they are, let's use cylindrical coordinates (p, theta, z), and center the pie on the origin, resting on top of the z=0 plane. We will integrate by disks. (Could go by shells, too, doesn't matter.)
You can set up the triple integral for
p dp d(theta) dz
where the limits are p=0 to radius
theta goes from 0 to 2 pi (all the way around)
z goes from 0 to 6
But what is the radius?
At z=0, p (rho, the radius) is 8. at z=6, p=12. You can derive the formula that p = 8 + 2z/3
so the limits of integration for p are 0 and 8 + 2z/3
Is that enough, can you do the integrals from there?
-
You can rotate a line around the x axis to get a pie. Locate (0,8) on the y axis, and (6,12).
Connect these two points and rotate this segment around the x axis.
This solid will be a truncated cone, the left end will be a circle with diameter 16, and the right end will be a circle with diameter 24.
The distance between them is the height, 6.
Find the volume by disk method.
The line would be y= (2/3)x+8 or (2x+24)/3
Integrate [ pi(2x+24)^2dx from 0 to 6= 608pi
:) R
Connect these two points and rotate this segment around the x axis.
This solid will be a truncated cone, the left end will be a circle with diameter 16, and the right end will be a circle with diameter 24.
The distance between them is the height, 6.
Find the volume by disk method.
The line would be y= (2/3)x+8 or (2x+24)/3
Integrate [ pi(2x+24)^2dx from 0 to 6= 608pi
:) R
-
the pie can be considered as a 6 cm. top slice of an inverted cone.
from simple geometry the depth of the total cone is 18 cm.
volume of total cone = 1/3 x pi x 12^2 x 18 = V...............solve for V
volume of 'total cone minus the slice' = 1/3 x pi x 8^2 x 12 = v.............solve for v
hence, volume of pie = (V - v ).................solve.
from simple geometry the depth of the total cone is 18 cm.
volume of total cone = 1/3 x pi x 12^2 x 18 = V...............solve for V
volume of 'total cone minus the slice' = 1/3 x pi x 8^2 x 12 = v.............solve for v
hence, volume of pie = (V - v ).................solve.