The volume of the region R bounded by the x-axis is: [tex]\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}[/tex]
What is the volume of the solid (R) on the X-axis?
If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.
From the given graph:
The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:
[tex]\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}[/tex]
here:
- (xâ, yâ) and (xâ, yâ) are two points on the straight line
Suppose we assign (xâ, yâ) = (0, 0) and (xâ, yâ) = (2, 8) Â from the graph, we have:
[tex]\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}[/tex]
y = 4x
Now, our region bounded by the three lines are:
Similarly, the change in polar coordinates is:
where;
- x² + y² = r²  and dA = rdrdθ
Therefore;
- rsinθ = 0  i.e.  r = 0 or θ = 0
- rcosθ = 2 i.e.  r  = 2/cosθ
- rsinθ = 4(rcosθ)  â tan θ = 4;  θ = tanâťÂš (4)
- â r = 0  to  r = 2/cosθ
-   θ = 0  to   θ = tanâťÂš (4)
Then:
[tex]\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}[/tex]
[tex]\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}[/tex]
Learn more about the determining the volume of solids bounded by region R here:
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