Math Problem Statement

solve the above question

Solve the above question.

Solution

Let’s analyze and solve the problems step by step. These problems involve integration in polar coordinates, geometry, Green's theorem, and exact differential forms.


5. (a) Find the area of the region common to the interiors of the cardioids r=1+cosθr = 1 + \cos\theta and r=1cosθr = 1 - \cos\theta.

  • Solution:
    The common region lies where the two cardioids intersect. The cardioids intersect when 1+cosθ=1cosθ1 + \cos\theta = 1 - \cos\theta, leading to cosθ=0\cos\theta = 0. This occurs at θ=π2\theta = \frac{\pi}{2} and θ=3π2\theta = \frac{3\pi}{2}.

The area of the common region is given by: A=12π23π2[(1+cosθ)2(1cosθ)2]dθA = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left[ (1+\cos\theta)^2 - (1-\cos\theta)^2 \right] \, d\theta Simplify: A=12π23π2[4cosθ]dθA = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left[ 4\cos\theta \right] \, d\theta A=2π23π2cosθdθA = 2 \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos\theta \, d\theta Evaluate: cosθdθ=sinθ\int \cos\theta \, d\theta = \sin\theta A=2[sinθ]π23π2=2(sin3π2sinπ2)A = 2 \left[ \sin\theta \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = 2 \left( \sin\frac{3\pi}{2} - \sin\frac{\pi}{2} \right) A=2(11)=4A = 2 \left( -1 - 1 \right) = -4 Since the area cannot be negative, the magnitude is 4units2\boxed{4 \, \text{units}^2}.


5. (b) Find the volume of the wedge cut from the cylinder x2+y2=1x^2 + y^2 = 1 by the plane z=yz = -y and z=0z = 0.

The cylinder x2+y2=1x^2 + y^2 = 1 is bounded between z=yz = -y and z=0z = 0. Using cylindrical coordinates, x=rcosθx = r\cos\theta, y=rsinθy = r\sin\theta, z=zz = z, and r=1r = 1, the limits are:

  • rr from 00 to 11,
  • θ\theta from 00 to 2π2\pi,
  • zz from rsinθ-r\sin\theta to 00.

Volume VV is: V=02π01rsinθ0rdzdrdθV = \int_0^{2\pi} \int_0^1 \int_{-r\sin\theta}^0 r \, dz \, dr \, d\theta Integrate with respect to zz: rsinθ0rdz=r[z]rsinθ0=r(0(rsinθ))=r2sinθ\int_{-r\sin\theta}^0 r \, dz = r \left[ z \right]_{-r\sin\theta}^0 = r \left( 0 - (-r\sin\theta) \right) = r^2 \sin\theta Integrate with respect to rr: 01r2sinθdr=sinθ01r2dr=sinθ[r33]01=sinθ3\int_0^1 r^2 \sin\theta \, dr = \sin\theta \int_0^1 r^2 \, dr = \sin\theta \left[ \frac{r^3}{3} \right]_0^1 = \frac{\sin\theta}{3} Integrate with respect to θ\theta: 02πsinθ3dθ=1302πsinθdθ=13[cosθ]02π=13(cos(2π)+cos(0))=0\int_0^{2\pi} \frac{\sin\theta}{3} \, d\theta = \frac{1}{3} \int_0^{2\pi} \sin\theta \, d\theta = \frac{1}{3} \left[ -\cos\theta \right]_0^{2\pi} = \frac{1}{3} \left( -\cos(2\pi) + \cos(0) \right) = 0 Hence, the volume is 0\boxed{0}.


Would you like me to proceed with the remaining parts or clarify any steps? Here's how you can deepen your understanding:

  1. Explain how polar integration works for finding areas.
  2. Explore why Green's theorem applies to problem 7(a).
  3. Discuss the physical interpretation of the cylindrical volume in 5(b).
  4. Review the exactness of differential forms for problem 7(b).
  5. Dive into spherical coordinates for volume computation in 6(b).

Tip: Break multi-dimensional integrals into small, simple steps and always check the bounds of integration!

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Math Problem Analysis

Mathematical Concepts

Polar coordinates and cardioids
Volumes using triple integration
Green's Theorem and line integrals
Exact differential forms

Formulas

Area in polar coordinates: A = 1/2 ∫[r(θ)]² dθ
Triple integral for volume: V = ∫∫∫ dz dr dθ
Green's Theorem: ∮C M dx + N dy = ∬R (∂N/∂x - ∂M/∂y) dA
Exact differential condition: ∂M/∂y = ∂N/∂x

Theorems

Green's Theorem
Properties of exact differential forms

Suitable Grade Level

Undergraduate-level calculus