Question

Use cylindrical coordinates to evaluate the integral where d is the solid bounded above by the plane z=2 and below by the surface 2z=x2+y2

Answer 1

Answer:

$\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \frac{32 \pi}{5}$

General Formulas and Concepts:

Calculus

Integration

• Integrals

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

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

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

Integration Property [Addition/Subtraction]:
$\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Multivariable Calculus

Cylindrical Coordinate Conversions:

• $\displaystyle x = r \cos \theta$
• $\displaystyle y = r \sin \theta$
• $\displaystyle z = z$
• $\displaystyle r^2 = x^2 + y^2$
• $\displaystyle \tan \theta = \frac{y}{x}$

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

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \text{Region} \ D \left \{ {{2z = x^2 + y^2} \atop {z = 2}} \right.$

$\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz$

Step 2: Find Volume Pt. 1

Find θ bound.

1. [Surface] Substitute in plane z:
   $\displaystyle 2(2) = x^2 + y^2$
2. Simplify:
   $\displaystyle x^2 + y^2 = 4$
3. [See 2nd Attachment] Graph
4. [Graph] Identify limits [Unit Circle]:
   $\displaystyle 0 \leq \theta \leq 2 \pi$

Find r bound.

1. [Surface] Substitute in plane z:
   $\displaystyle 2(2) = x^2 + y^2$
2. Substitute in cylindrical conversions:
   $\displaystyle 2(2) = r^2$
3. Simplify:
   $\displaystyle r = \pm 2$
4. [r] Identify:
   $\displaystyle r = 2$
5. Define limits:
   $\displaystyle 0 \leq r \leq 2$

Find z bound.

1. [Surface] Substitute in cylindrical conversions:
   $\displaystyle 2z = r^2$
2. Rewrite:
   $\displaystyle z =\frac{r^2}{2}$
3. Define limits:
   $\displaystyle \frac{r^2}{2} \leq z \leq 2$

Step 3: Find Volume Pt. 2

1. [Integrals] Substitute in cylindrical conversions:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \iiint_D {z \big( r^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz$
2. [Integrals] Simplify:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \iiint_D {\frac{z}{r}} \, dx \, dy \, dz$
3. [Integrals] Convert [Volume Formula - Cylindrical Coordinates]:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \iiint_T {\frac{zr}{r}} \, dz \, dr \, d\theta$
4. [Integrals] Simplify:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \iiint_T {z} \, dz \, dr \, d\theta$
5. [Integrals] Substitute in region T:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \int\limits^{2 \pi}_0 \int\limits^2_0 \int\limits^2_{\frac{r^2}{2}} {z} \, dz \, dr \, d\theta$
6. [dz Integral] Integrate [Integration Rules and Properties]:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \int\limits^{2 \pi}_0 \int\limits^2_0 {\frac{z^2}{2} \bigg| \limits^{z = 2}_{z = \frac{r^2}{2}}} \, dr \, d\theta$
7. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \int\limits^{2 \pi}_0 \int\limits^2_0 {\bigg( 2 - \frac{r^4}{8} \bigg)} \, dr \, d\theta$
8. [dr Integral] Integrate [Integration Rules and Properties]:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \int\limits^{2 \pi}_0 {\bigg( 2r - \frac{r^5}{40} \bigg) \bigg| \limits^{r = 2}_{r = 0}} \, d\theta$
9. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \int\limits^{2 \pi}_0 {\frac{16}{5}} \, d\theta$
10. [dθ Integral] Integrate [Integration Rules and Properties]:
    $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \frac{16}{5} \theta \bigg| \limits^{\theta = 2 \pi}_{\theta = 0}$
11. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
    $\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \frac{32 \pi}{5}$

∴ the integral bound by region D is equal to  $\displaystyle \bold{\frac{32 \pi}{5}}$.

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Learn more about cylindrical coordinates: https://brainly.com/question/24004264

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

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Topic: Multivariable Calculus

Unit: Triple Integrals Applications