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2 answers:
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First of all, let's recall the area of a triangle, knowing its base (b) and height (h):
The exercise is showing you that, if you inscribe a polynomial with more and more side, the area of the polynomial will approximate the area of the circle better and better (you can see youself that the polygon is "filling" the circle more and more as the number of sides increase).
Now, the second column tells you the area of each of the triangles the polygon is split into. So, we can see that the first polygon is split into 3 triangles, each of them having base 1.73 and height 0.5.
So, the area of each triangle is
There are three of these triangles, so the area of the whole polygon is
In the second case, you have six triangles, each with base 1 and height 0.87. So, the whole area is
Finally, in the last case you have 8 triangles, each with base 0.77 and height 0.92. So, the whole area is
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