Let R be the triangular region in the first quadrant with vertices at points (0,0), (a,0), and (0,b), where a and b are positive constants. Write dow the volume of the solid generated when region R is revolved about the x-axis?

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

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

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step-by-step explanation: