Use cylindrical coordinates to evaluate the triple integral dV= function equation editor , where EE is the region that lies inside the cylinder x2+y2=4x2+y2=4 and between the planes z=2z=2 and z=7z=7.

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Answer:

[tex]\displaystyle \iiint_E \, dV = 20 \pi[/tex]

General Formulas and Concepts:

Calculus

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Multivariable Calculus

Cylindrical Coordinate Conversions:

  • [tex]\displaystyle x = r \cos \theta[/tex]
  • [tex]\displaystyle y = r \sin \theta[/tex]
  • [tex]\displaystyle z = z[/tex]
  • [tex]\displaystyle r^2 = x^2 + y^2[/tex]
  • [tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]

Volume Formula [Cylindrical Coordinates]:
[tex]\displaystyle V = \iiint_T \, dV \rightarrow V = \iiint_T {r} \, dz \, dr \, d\theta[/tex]

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \text{Region} \ E \left\{ \begin{array}{ccc} \text{Cylinder} \ x^2 + y^2 = 4 \\ \text{Plane} \ z = 2 \\ \text{Plane} \ z = 7 \end{array}[/tex]

Step 2: Integrate Pt. 1

Find z bounds.

  1. [Given] Define:
    [tex]\displaystyle 2 \leq z \leq 7[/tex]

Find r bounds.

  1. [Cylinder] Substitute in cylindrical conversions:
    [tex]\displaystyle x^2 + y^2 = 4 \rightarrow r^2 = 4[/tex]
  2. Simplify:
    [tex]\displaystyle r = \pm 2[/tex]
  3. [r] Identify:
    [tex]\displaystyle r = 2[/tex]
  4. Define limits:
    [tex]\displaystyle 0 \leq r \leq 2[/tex]

Find θ bounds.

  1. [Cylinder] Graph [See 2nd Attachment]
  2. Identify limits:
    [tex]\displaystyle 0 \leq \theta \leq 2 \pi[/tex]

Step 3: Integrate Pt. 2

  1. [Integrals] Convert [Volume Formula - Cylindrical Coordinates]:
    [tex]\displaystyle \iiint_T \, dV = \iiint_T {r} \, dz \, dr \, d\theta[/tex]
  2. [Integrals] Substitute in region T:
    [tex]\displaystyle \iiint_T \, dV = \int\limits^{2 \pi}_0 \int\limits^{2}_0 \int\limits^7_2 {r} \, dz \, dr \, d\theta[/tex]
  3. [dz Integral] Apply Integration Rule [Reverse Power Rule]:
    [tex]\displaystyle \iiint_T \, dV = \int\limits^{2 \pi}_0 \int\limits^{2}_0 {rz \bigg| \limits^{z = 7}_{z = 2}} \, dr \, d\theta[/tex]
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus]:
    [tex]\displaystyle \iiint_T \, dV = \int\limits^{2 \pi}_0 \int\limits^{2}_0 {5r} \, dr \, d\theta[/tex]
  5. [Integrals] Rewrite [Integration Property - Multiplied Constant]:             [tex]\displaystyle \iiint_T \, dV = 5 \int\limits^{2 \pi}_0 \int\limits^{2}_0 {r} \, dr \, d\theta[/tex]
  6. [dr Integral] Apply Integration Rule [Reverse Power Rule]:
    [tex]\displaystyle \iiint_T \, dV = 5 \int\limits^{2 \pi}_0 {\frac{r^2}{2} \bigg| \limits^{r = 2}_{r = 0}} \, d\theta[/tex]
  7. Evaluate [Integration Rule - Fundamental Theorem of Calculus]:
    [tex]\displaystyle \iiint_T \, dV = 5 \int\limits^{2 \pi}_0 {2} \, d\theta[/tex]
  8. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    [tex]\displaystyle \iiint_T \, dV = 10 \int\limits^{2 \pi}_0 {} \, d\theta[/tex]
  9. [Integral] Apply Integration Rule [Reverse Power Rule]:
    [tex]\displaystyle \iiint_T \, dV = 10 \theta \bigg| \limits^{\theta = 2 \pi}_{\theta = 0}[/tex]
  10. Evaluate [Integration Rule - Fundamental Theorem of Calculus]:
    [tex]\displaystyle \iiint_T \, dV = 10(2 \pi)[/tex]
  11. Simplify:
    [tex]\displaystyle \iiint_E \, dV = 20 \pi[/tex]

∴ the integral bound by region D is equal to 20π.

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Learn more about cylindrical coordinates: https://brainly.com/question/6177409

Learn more about multivariable calculus: https://brainly.com/question/17203772

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Topic: Multivariable Calculus

Unit: Triple Integrals Applications

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