{"id":31197,"date":"2021-04-08T14:53:01","date_gmt":"2021-04-08T13:53:01","guid":{"rendered":"http:\/\/kmr.dialectica.se\/wp\/?page_id=31197"},"modified":"2022-03-11T16:04:42","modified_gmt":"2022-03-11T15:04:42","slug":"examples","status":"publish","type":"page","link":"https:\/\/kmr.placify.me\/wp\/research\/math-rehab\/learning-object-repository\/calculus\/calculus-of-several-real-variables\/integration-of-functions-of-several-real-variables\/substitution-in-integrals\/separating-the-variables\/examples\/","title":{"rendered":"Integration examples"},"content":{"rendered":"

This is a sub-page of our page on Separating the variables<\/a><\/p>\n

\/\/\/\/\/\/\/<\/p>\n

Related KMR pages<\/strong><\/p>\n

\u2022 …<\/p>\n

\/\/\/\/\/\/\/<\/p>\n

Other relevant sources of information<\/strong><\/p>\n

\u2022 …<\/p>\n

\/\/\/\/\/\/\/<\/p>\n

\/\/\/\/\/\/\/ Problem 1<\/strong>:<\/p>\n

Formulation<\/strong>: Calculate the volume of the ellipsoid \\, \\dfrac{x^2}{a^2} + \\dfrac{y^2}{b^2} + \\dfrac{z^2}{c^2} \\leq 1 \\, <\/span><\/p>\n

Solution<\/strong>: \\;\\;\\;\\;\\; <\/span> If we make the substitution \\, \\begin{Bmatrix} \\begin{smallmatrix} x(t, \\theta, \\phi) \\, = \\, a \\, t \\sin\\theta \\cos\\phi \\\\ y(t, \\theta, \\phi) \\, = \\, b \\, t \\sin\\theta \\sin\\phi \\\\ \\; z(t, \\theta, \\phi) \\, = \\, c \\, t \\cos\\theta \\;\\;\\;\\;\\;\\;\\;\\, \\end{smallmatrix} \\end{Bmatrix} \\, <\/span><\/p>\n

the differential<\/a> transformation of \\, (x, y, z) \\, <\/span> with respect to \\, (t, \\theta, \\phi) \\, <\/span> is given by<\/p>\n

\\, \\begin{pmatrix} \\begin{smallmatrix} dx \\\\ dy \\\\ dz \\end{smallmatrix} \\end{pmatrix} = \\frac {\\partial(x, y, z)}{\\partial(t, \\theta, \\phi)} \\begin{pmatrix} \\begin{smallmatrix} dt \\\\ d\\theta \\\\ d\\phi \\end{smallmatrix} \\end{pmatrix} <\/span>,<\/p>\n

where the Jacobian matrix<\/a> \\, \\frac {\\partial(x, y, z)}{\\partial(t, \\theta, \\phi)} = \\begin{pmatrix} \\begin{smallmatrix} a \\sin\\theta \\cos\\phi & a \\, t \\cos\\theta \\cos\\phi & -a \\, t \\sin\\theta \\sin\\phi \\\\ b \\sin\\theta \\sin\\phi & b \\, t \\cos\\theta \\cos\\phi & \\;\\; b \\, t \\sin\\theta \\cos\\phi \\\\ c \\cos\\phi & -c \\, t \\sin\\phi & 0 \\end{smallmatrix} \\end{pmatrix} <\/span>,<\/p>\n

and the Jacobian determinant \\vert \\frac {\\partial(x, y, z)}{\\partial(t, \\theta, \\phi)} \\vert = \\begin{smallmatrix} a b c \\, t^2 \\sin\\theta \\end{smallmatrix} <\/span>.<\/p>\n

Assembling these pieces we get<\/p>\n \\, \\iiint\\limits_{\\frac{x^2}{a^2} + \\frac{y^2}{b^2} + \\frac{z^2}{c^2} \\leq 1} dx dy dz \\, = \\, \\begin{Bmatrix} \\begin{smallmatrix} \\text{substitution:} \\\\ x \\, = \\, a \\, t \\sin\\theta \\cos\\phi \\\\ y \\, = \\, b \\, t \\sin\\theta \\sin\\phi \\\\ \\; z \\, = \\, c \\, t \\cos\\theta \\;\\;\\;\\;\\;\\;\\;\\, \\\\ dx dy dz \\, = \\, a b c \\, t^2 \\sin\\theta \\, dt d\\theta d\\phi \\end{smallmatrix} \\end{Bmatrix} \\, = <\/span>\n \\, = \\iiint\\limits_{ \\begin{smallmatrix} 0\u2264t\u22641, \\\\ 0\u2264\\theta\u2264\\pi, \\\\ 0\u2264\\phi\u2264 2 \\pi \\end{smallmatrix} } a b c \\, t^2 \\sin\\theta \\, dt d\\theta d\\phi \\, = \\, <\/span>\n \\, = \\, a b c \\int\\limits_{0}^{1} t^2 dt \\, \\int\\limits_{0}^{\\pi} \\sin\\theta d\\theta \\, \\int\\limits_{0}^{2 \\pi} d \\phi \\, = \\, a b c \\, [\\frac{1}{3} t^3]_0^1 \\, [- \\cos\\theta]_0^{\\pi} \\, [ \\phi ]_0^{2 \\pi} \\, = <\/span>\n

\\, = a b c \\cdot \\frac{1}{3} \\cdot 2 \\cdot 2 \\pi \\, = \\, \\frac{ 4 \\pi}{3} a b c <\/span>.<\/p>\n

REFLECTION<\/strong>: The substitution of variables transforms the ellipsoid<\/a> in \\, (x, y, z) <\/span>-space into a rectangular box<\/a> in \\, (t, \\theta, \\phi) <\/span>-space. In the latter space the effects of the variables on the integral are independent of each other, which enables the integral to be separated into a product of three independent one-dimensional integrals.<\/p>\n

Note also that if \\, a = b = c = r \\, <\/span> then the volume of the ellipsoid becomes \\, \\frac {4\\pi}{3} r^3 \\, <\/span>,
\nwhich is equal to the volume of a sphere with radius \\, r \\, <\/span> (as it should be).<\/p>\n

\/\/\/
\n \\, \\begin{pmatrix} \\begin{smallmatrix} dx \\\\ dy \\\\ dz \\end{smallmatrix} \\end{pmatrix} = \\begin{pmatrix} \\begin{smallmatrix} a \\sin\\theta \\cos\\phi & a \\, t \\cos\\theta \\cos\\phi & -a \\, t \\sin\\theta \\sin\\phi \\\\ b \\sin\\theta \\sin\\phi & b \\, t \\cos\\theta \\cos\\phi & b \\, t \\sin\\theta \\cos\\phi \\\\ c \\cos\\phi & -c \\, t \\sin\\phi & 0 \\end{smallmatrix} \\end{pmatrix} \\begin{pmatrix} \\begin{smallmatrix} dt \\\\ d\\theta \\\\ d\\phi \\end{smallmatrix} \\end{pmatrix} <\/span>.
\n\/\/\/<\/p>\n

\/\/\/\/\/\/\/ Problem 2<\/strong>:<\/p>\n

Formulation<\/strong>: Calculate the integral \\, \\iiint z \\sqrt{x^2+y^2} dx dy dz \\, <\/span>
\nover the body enclosed by the blue paraboloid \\, z = x^2+y^2 \\, <\/span>
\nand the yellow plane \\, z = 1 <\/span>.<\/p>\n

\"z=x^2+y^2,<\/a><\/p>\n

Solution<\/strong>:<\/p>\n \\, \\iiint\\limits_{x^2+y^2\u2264z\u22641} z \\sqrt{x^2+y^2} dx dy dz = <\/span>\n \\, = \\iint\\limits_{x^2+y^2\u22641} \\sqrt{x^2+y^2} dx dy \\, \\int\\limits_{x^2+y^2}^{1} z dz = <\/span>\n \\, = \\iint\\limits_{x^2+y^2\u22641} \\sqrt{x^2+y^2} dx dy \\, { [ \\,\\frac{1}{2} z^2 \\, ] }_{x^2+y^2}^{1} = <\/span>\n \\, = \\iint\\limits_{x^2+y^2\u22641} \\sqrt{x^2+y^2} \\, \\frac{1}{2} (1 - (x^2+y^2)^2) dx dy = \\begin{Bmatrix} \\begin{smallmatrix} \\text{polar coordinates:} \\\\ x(r, \\theta) \\, = \\, r \\cos\\theta \\\\ y(r, \\theta) \\, = \\, r \\sin\\theta \\\\ \\vert \\frac {\\partial(x, y)} {\\partial(r, \\theta)} \\vert \\, = \\, r \\\\ dx dy = r dr d\\theta \\end{smallmatrix} \\end{Bmatrix} = <\/span>\n \\, = \\frac{1}{2} \\int\\limits_{0}^{2\\pi} d\\theta \\, \\int\\limits_{0}^{1} (1 - r^4) r \\cdot r dr = \\frac{1}{2} \\, 2\\pi \\, \\int\\limits_{0}^{1} (r^2 - r^6) dr = \\, <\/span>\n

\\, = \\pi \\, \\left[ \\, \\dfrac{1}{3} r^3 - \\dfrac{1}{7} r^7 \\, \\right]_{0}^{1} = \\pi \\left( \\dfrac{1}{3} - \\dfrac{1}{7} \\right) = \\dfrac{4\\pi}{21} <\/span>.<\/p>\n

REFLECTION<\/strong>: In this integral the variable \\, z \\, <\/span> is in fact operating independently from the variables \\, x \\, <\/span> and \\, y <\/span>. It is only the lower limit of the \\, z <\/span>-integral that depends on \\, x \\, <\/span> and \\, y <\/span>. This is why we can get away with integrating first in the \\, z <\/span>-direction and then substitute into polar coordinates in order to separate the \\, x \\, <\/span> and \\, y <\/span> parts of the integral.<\/p>\n

\/\/\/\/\/\/\/<\/p>\n","protected":false},"excerpt":{"rendered":"

This is a sub-page of our page on Separating the variables \/\/\/\/\/\/\/ Related KMR pages \u2022 … \/\/\/\/\/\/\/ Other relevant sources of information \u2022 … \/\/\/\/\/\/\/ \/\/\/\/\/\/\/ Problem 1: Formulation: Calculate the volume of the ellipsoid Solution: If we make the substitution the differential transformation of with respect to is given by , where the … Continue reading Integration examples<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":30262,"menu_order":0,"comment_status":"open","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":["post-31197","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/pages\/31197","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/comments?post=31197"}],"version-history":[{"count":2,"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/pages\/31197\/revisions"}],"predecessor-version":[{"id":33762,"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/pages\/31197\/revisions\/33762"}],"up":[{"embeddable":true,"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/pages\/30262"}],"wp:attachment":[{"href":"https:\/\/kmr.placify.me\/wp\/wp-json\/wp\/v2\/media?parent=31197"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}