Help I really need the answer to this.
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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
The graph of the equation is a linear graph and is as represented in the attached image.
<h3>Graph of the equation:</h3>According to the question;
- The profit y (in dollars) for a business from selling x games is represented by; y = 33x.
In essence, the graph of the equation is a straight line graph with slope, m = 33.
The graph of the equation is therefore as represented in the attached image.
Read more on straight line graph;
