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:
- (xā, yā) and (xā, yā) are two points on the straight line
Suppose we assign (xā, yā) = (0, 0) and (xā, yā) = (2, 8) Ā 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;
- xĀ² + yĀ² = rĀ² Ā and dA = rdrdĪø
Therefore;
- rsinĪø = 0 Ā i.e. Ā r = 0 or Īø = 0
- rcosĪø = 2 i.e. Ā r Ā = 2/cosĪø
- rsinĪø = 4(rcosĪø) Ā ā tan Īø = 4; Ā Īø = tanā»Ā¹ (4)
- ā r = 0 Ā to Ā r = 2/cosĪø
- Ā Ā Īø = 0 Ā to Ā Ā Īø = tanā»Ā¹ (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]
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