5. Triple Integrals Solutions

5. Triple Integrals Solutions - 5 Triple Integrals 5A...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
5. Triple Integrals 5A. Triple integrals in rectangular and cylindrical coordinates 1 Middle: iy + iy2 + yz ] = 1 + z - (-1) = 1 + 2z Outer: z + z2] 2 = 6 y=-1 0 X Y dz dx dy Inner: xy2z2] o = x3y4 Jd2 Jdfi JdxY b) ~ X Y ~ Z Middle: ,x 2 1 4 y 4 ] f i = iy6 Outer: kY7] = y. 0 5A-2 dz dy dx (ii) Jdl Jdl-' Jdl dxdzdy (iii)JdlJdlJdl-'dydXdz c) In cylindrical coordinates, with the polar coordinates r and 8 in xz-plane, we get 11, dy dr de = Jdr" Jdl Jd2 dy dr do d) The sphere has equation x2 + y2 + z2 = 2, or r2 + z2 = 2 in cylindrical coordinates. The cone has equation z2 = r2, or z = r. The circle in which they intersect has a radius r found by solving the two equations z = r and z2+r2 = 2 simultaneously; .Y eliminating z we get r2 = 1, so r = 1. Putting it all together, we get \ : I cross-sectionview 5A-3 By symmetry, i = y = 2, so it suffices to calculate just one of these, say 2. We have 1-x 1-z-y z-moment = z d v = z dz dy dx z= 1-X-y Jdl Jd Jd 1-x-y 1-2 1nner: iz2] = ; ( l - ~ - y ) ~ M i d d l e : - i ( l - ~ - y ) ~ ] ~ = i ( l - ~ ) ~ 0 1 Outer: - 1 - x 4 ] = I - z moment. - ,, 0 v= 1-x mass of D = volume of D = i(base)(height) = i - ; . 1 = i . Therefore 2 = &/+ = i; this is also Z and g, by symmetry. 5A-4 Placing the cone as shown, its equation in cylindrical coordinates is z = r and the density is given by 6 = r. By the geometry, its projection onto the xy-plane is the interior R of the origin-centered circle of radius h. vertical cross-section
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
TRIPLE INTEGRALS a) Mass of solid D = / ~ ~ ~ 6 d V = l ~ ~ l ~ ~ l ~ ~ . r d z d r d 6 2 ~ h 4 Inner: (h - r)r2; Middle: - - - = . Outer: - 3 hr3 4 , 12' 12 r41h b) By symmetry, the center of mass is on the z-axis, so we only have to compute its z-coordinate, 5. z-moment of D = / / L z 6 d v = ~ 2 r ~ h J l h z T ~ T d z d T d ~ h 1 Inner: iz2r2] = i(h2r2 - r4) Middle: g) = -h5 . - 2 2 A (h2: 2 15 - T , 5A-5 Position S so that its base is in the xy-plane and its diagonal D lies along the x-axis (the y-axis would do equally well). The octants divide S into four tetrahedra, which by symmetry have the same moment of inertia about the x-axis; we calculate the one in the first octant. (The picture looks like that for 5A-3, except the height is 2.)
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern