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Aneli [31]
2 years ago
13

Is y/x a positive value together

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
3 0

Answer:

yes

Step-by-step explanation:

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Kobotan [32]

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}

<h3>What is the volume of the solid (R) on the X-axis?</h3>

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:

brainly.com/question/14393123

#SPJ1

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