Question

Need your help now please; the assignment is due tonight and this is the last problem I am having trouble with. The red box is the value I keep getting wrong

The region R is bounded by the x-axis, the straight line in the graph and the vertical line x=2.

Answer 1

The volume of the region R bounded by the x-axis is: $\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}$

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:

$\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}$

here:

• (x₁, y₁) and (x₂, y₂) are two points on the straight line

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, we have:

$\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}$

y = 4x

Now, our region bounded by the three lines are:

• y = 0
• x = 2
• y = 4x

Similarly, the change in polar coordinates is:

• x = rcosθ,
• y = rsinθ

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)  ⇒ tan θ = 4;  θ = tan⁻¹ (4)

• ⇒ r = 0   to   r = 2/cosθ
•    θ = 0  to    θ = tan⁻¹ (4)

Then:

$\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}$

$\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}$

Learn more about the determining the volume of solids bounded by region R here:

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