Question

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?

Answer 1

Answer:

volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( $\frac{-b}{a}$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 = $\frac{-b}{a}$x + ba/a

y = $\frac{-b}{a}$x + b

so R is bounded by  y = $\frac{-b}{a}$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 ( $\frac{-b}{a}$x + b )² dx

V = π ₀∫^a ( $\frac{-b}{a}$x + b )² dx

Therefore,  volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( $\frac{-b}{a}$x + b )² dx

Answer 2

The volume of the solid generated when region R is revolved about the x-axis is $\frac{b^{5} }{3a^{2} }$.

Equation of the line AB

$y-0 = \frac{b-0}{0-a} (x-a)$

$y = \frac{-b}{a} (x-a)$

What is the volume generated when a curve f(x) is generated about the x-axis?

The volume generated when a curve f(x) is rotated about the x-axis in x∈(c,d) is given by:

$V=\int\limits^d_c {y^2} \, dx$

So, the volume generated when line AB is rotated about the x-axis will be $V=\int\limits^b_0 ({\frac{-b}{a} (x-a))^2} \, dx$

$V=\frac{b^{5} }{3a^{2} }$

Therefore, the volume of the solid generated when region R is revolved about the x-axis is $\frac{b^{5} }{3a^{2} }$.

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