Answer:
volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( 
x + b )² dx
Step-by-step explanation:
Given the data in the question and as illustrated in the image below;
R is in the region first quadrant with vertices; 0(0,0), A(a,0) and B(0,b)
from the image;
the equation of AB will be;
y-b / b-0 = x-0 / 0-a
(y-b)(0-a) = (b-0)(x-0)
0 - ay -0 + ba = bx - 0 - 0 + 0
-ay + ba = bx
ay = -bx + ba
divide through by a
y = x + ba/a
y = x + b
so R is bounded by y = x + b and y =0, 0 ≤ x ≤ a
The volume of the solid revolving R about x axis is;
dv = Area × thickness
= π( Radius)² dx
= π ( x + b )² dx
V = π ₀∫^a ( x + b )² dx
Therefore, volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( x + b )² dx
Answer:
1/3
Step-by-step explanation:
Answer:
(5 - y) ^3 = 125 - 75y + 15y^2 - y^3
Step-by-step explanation:
Binomial expression
1
1. 1
1. 2. 1
1. 3. 3. 1 --------power of 3
( 5 - y) ^3
( 5 - y) (5 - y) (5 - y)
( a + b) ^3 = a^3 + 3a^2b + 3ab^2 + b^3
a = 5
b = -y
( 5 - y) ^2 = ( 5 - y) (5 - y)
= 5( 5 - y) - y(5 - y)
= 25 - 5y - 5y + y^2
=(25-10y+y^2)
( 25 - 10y + y^2)( 5 - y)
= 5(25 - 10y + y^2) - y( 25 - 10y + y^2)
= 125 - 50y + 5y^2 - 25y + 10y^2 - y^3
Collect the like terms
= 125 - 50y - 25y + 5y^2 + 10y^2 - y^3
= 125 - 75y + 15y^2 - y^3
Volume = <span>πd^2h/4
245 = </span><span>π x d^2 x 5/4
d^2 = (245 x 4)/5</span><span>π = 62.39
d = sqrt(62.39) = 7.9 units
</span>
