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:[tex]\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \frac{32 \pi}{5}[/tex]General Formulas and Concepts:CalculusIntegrationIntegralsIntegration 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]Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]Multivariable CalculusCylindrical 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: DefineIdentify.[tex]\displaystyle \text{Region} \ D \left \{ {{2z = x^2 + y^2} \atop {z = 2}} \right.[/tex][tex]\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz[/tex]Step 2: Find Volume Pt. 1Find θ bound.[Surface] Substitute in plane z:
[tex]\displaystyle 2(2) = x^2 + y^2[/tex]Simplify:
[tex]\displaystyle x^2 + y^2 = 4[/tex][See 2nd Attachment] Graph[Graph] Identify limits [Unit Circle]:
[tex]\displaystyle 0 \leq \theta \leq 2 \pi[/tex]Find r bound.[Surface] Substitute in plane z:
[tex]\displaystyle 2(2) = x^2 + y^2[/tex]Substitute in cylindrical conversions:
[tex]\displaystyle 2(2) = r^2[/tex]Simplify:
[tex]\displaystyle r = \pm 2[/tex][r] Identify:
[tex]\displaystyle r = 2[/tex]Define limits:
[tex]\displaystyle 0 \leq r \leq 2[/tex]Find z bound.[Surface] Substitute in cylindrical conversions:
[tex]\displaystyle 2z = r^2[/tex]Rewrite:
[tex]\displaystyle z =\frac{r^2}{2}[/tex]Define limits:
[tex]\displaystyle \frac{r^2}{2} \leq z \leq 2[/tex]Step 3: Find Volume Pt. 2[Integrals] Substitute in cylindrical conversions:
[tex]\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[/tex][Integrals] Simplify:
[tex]\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[/tex][Integrals] Convert [Volume Formula - Cylindrical Coordinates]:
[tex]\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[/tex][Integrals] Simplify:
[tex]\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \iiint_T {z} \, dz \, dr \, d\theta[/tex][Integrals] Substitute in region T:
[tex]\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[/tex][dz Integral] Integrate [Integration Rules and Properties]:
[tex]\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[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\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[/tex][dr Integral] Integrate [Integration Rules and Properties]:
[tex]\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[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\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[/tex][dθ Integral] Integrate [Integration Rules and Properties]:
[tex]\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}[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \iiint_D {z \big( x^2 + y^2 \big)^\big{\frac{-1}{2}}} \, dx \, dy \, dz = \frac{32 \pi}{5}[/tex]∴ the integral bound by region D is equal to  [tex]\displaystyle \bold{\frac{32 \pi}{5}}[/tex].---Learn more about cylindrical coordinates: more about multivariable calculus: : Multivariable CalculusUnit: Triple Integrals Applications