Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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.

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 class=

Sagot :

ajeigbeibraheem

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:

  • (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:

[tex]\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}[/tex]

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:

[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:

https://brainly.com/question/14393123

#SPJ1

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.