Set up the triple integral of a function g(x,y,z) over the region bounded below by the cone z=...

Question:

Set up the triple integral of a function g(x,y,z) over the region bounded below by the cone {eq}z= (x^2+y^2)^(1/2) {/eq} and above by the sphere {eq}x^2+y^2+z^2= 4 {/eq}, in cartesian, cylindrical and spherical coordinates.

Triple Integrals:

{eq}\\ {/eq}

Integrals defined over regions in space have to be integrated thrice. Triple integrals are vastly used in computing mass and volumes of solids.

Depending upon the region of integration, a suitable coordinate system: Cartesian, Spherical or Cylindrical, can be selected for evaluating the integral.

Answer and Explanation:

{eq}\\ {/eq}

The function {eq}g(x,y,z) {/eq} is to be integrated over a region bounded below by the cone: {eq}z = \sqrt{x^2+y^2} {/eq} and above by a sphere: {eq}x^2+y^2+z^2 = 4 {/eq}.

{eq}\\ {/eq}

a)

The cone and sphere intersect in a disk which is given by:

{eq}\begin{align*} \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2} \implies x^2+y^2 = 2 \end{align*} {/eq}

It is disk centered at the origin with a radius {eq}\sqrt{2} {/eq} and lies in the plane: {eq}z = \sqrt{x^2+y^2} = \sqrt{2} {/eq}.

In the region, x and y vary over the disk and z varies from {eq}z=\sqrt{x^2+y^2} {/eq} to {eq}z=\sqrt{4-x^2-y^2} {/eq}.

Over the disk: {eq}-\sqrt{2} \leq x \leq \sqrt{2}, \ -\sqrt{2-x^2} \leq y \leq \sqrt{2-x^2} {/eq}.

Therefore, the integral is set up in the Cartesian coordinates as:

{eq}\begin{align*} I &= \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}} g(x,y,z) \ \mathrm{d}z\mathrm{d}y\mathrm{d}x \end{align*} {/eq}

{eq}\\ {/eq}

b)

In the cylindrical coordinates {eq}(r, \theta, z) {/eq}, for the region of integration:

r and {eq}\theta {/eq} vary over the disk while z varies from {eq}z= \underbrace{r}_{\sqrt{x^2+y^2}} {/eq} to {eq}z=\underbrace{\sqrt{4-r^2}}_{\sqrt{4-(x^2+y^2)}} {/eq}.

Over the disk: {eq}0 \leq r \leq \sqrt{2}, \ 0 \leq \theta \leq 2\pi {/eq}.

For the rectangular to cylindrical transformation, the Jacobian results in: {eq}\mathrm{d}x\mathrm{d}y\mathrm{d}z = r\mathrm{d}z\mathrm{d}r\mathrm{d}\theta {/eq}

Therefore, the integral is set up in the cylindrical coordinates in the following way:

{eq}\begin{align*} I = \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{r}^{\sqrt{4-r^2}} g(r, \theta, z) \ r\mathrm{d}z\mathrm{d}r\mathrm{d}\theta \end{align*} {/eq}

{eq}\\ {/eq}

c)

In the spherical coordinates {eq}(\rho, \theta, \phi) {/eq}, for the region:

{eq}\rho = \sqrt{x^2+y^2+z^2} = 2 {/eq}, therefore {eq}0 \leq \rho \leq 2 {/eq}.

The top of the sphere is at {eq}\phi = 0 {/eq} and the cone: {eq}\sqrt{x^2+y^2} {/eq} is given by {eq}\phi = \pi/4 {/eq}, therefore: {eq}0 \leq \phi \leq \pi/4 {/eq}.

{eq}\theta {/eq} over the region varies from 0 to {eq}2\pi {/eq}.

For the rectangular to spherical transformation, the Jacobian results in: {eq}\mathrm{d}x\mathrm{d}y\mathrm{d}z = \rho^2 \sin{\phi} \mathrm{d}\phi\mathrm{d}r\mathrm{d}\theta {/eq}

Therefore, the integral is set up in the spherical coordinates in the following way:

{eq}\begin{align*} I = \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{\pi/4} g(\rho, \theta, \phi) \ \rho^2 \sin{\phi} \mathrm{d}\phi\mathrm{d}r\mathrm{d}\theta \end{align*} {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

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