# Recitation 12

Section 12.5 Problem 12: Evaluate the triple integral. \iiint_E xy dV, where E is bounded by the parabolic cylinders y=x^2 and x=y^2 and the planes z=0 and z=x+y.

Comment: Denote the region on xy-plane bounded by y=x^2 and x=y^2 by L. We obtain \iiint_E xy dV = \iint_L xy(x+y)dA = \int_0^1\int_{x^2}^{\sqrt{x}}xy(x+y)dydx = \frac{3}{28}.

Section 12.5 Problem 18: Use a triple integral to find the volume of the given solid. The solid enclosed by the paraboloids y = x^2 + z^2 and y=8-x^2-z^2.

Solution: The projection of the solid on the xz-plane is a disc with radius 2 centered at the origin, denoted by D. The volume is given by \iint_D 8-x^2-z^2-(x^2+z^2) dA. Under the polar coordinate of the xz-plane, this integral becomes \int_0^{2\pi}\int_0^2 (8-2r^2)r drd\theta = 16\pi.

Section 12.5 Problem 35: Evaluate the triple integral using only geometric interpretation and symmetry. \iiint_C (4 + 5x^2yz^2) dV, where C is the cylindrical region x^2 +y^2 \leq 4,-2\leq z\leq 2.

Solution: By symmetry \iiint_C (4 + 5x^2yz^2) dV = \iiint_C 4 dV. By geometric interpretation, this is 4 times of the volume of the cylindrical region whose volume is \pi\times 2^2\times 4=16\pi. Hence the answer is 64\pi.

Section 12.6 Problem 16: Sketch the solid whose volume is given by the integral and evaluate the integral. \int_0^2\int_0^{2\pi}\int_0^r r dz d\theta dr.

Answer: This is part of the region inside the cylinder x^2+y^2=2^2 that lies above the plane z=0 and below the cone z=\sqrt{x^2+y^2}.

Section 12.6 Problem 18: Use cylindrical coordinates. Evaluate \iiint_E z dV, where E is enclosed by the paraboloid z=x^2 +y^2 and the plane z=4.

Solution: The projection of the solid on the xy-plane is the disc centered at the origin with radius 2, denoted by D. Then \iiint_E z dV = \iint_D\int_{x^2+y^2}^4 z dz dA. Under the polar coordinate, this is equal to \int_0^{2\pi}\int_0^2\int_{r^2}^4 z dz r dr d\theta = \frac{64}{3}\pi.

Section 12.6 Problem 19: Use cylindrical coordinates. Evaluate \iiint_E (x + y + z) dV, where E is the solid in the first octant that lies under the paraboloid z = 4 - x^2 - y^2.

Solution: The projection of the solid onto the xy-plane is the part of disc with radius 2 centered at the origin in the first quadrant. Under the cylindrical coordinates, \iiint_E (x + y + z) dV = \int_0^{\pi/2}\int_0^2\int_0^{4-r^2} (r\cos\theta+r\sin\theta+z)dz r dr d\theta=\frac{208}{15}.

Section 12.7 Problem 15: A solid lies above the cone z = \sqrt{x^2 + y^2} and below the sphere x^2 + y^2 + z^2 = z. Write a description of the solid in terms of inequalities involving spherical coordinates.

Solution: Under the spherical coordinates, the cone is described by \phi = \pi / 4 and the sphere by \rho=\cos\phi. Therefore, the solid is described by 0\leq\theta\leq 2\pi, 0\leq \phi\leq \pi/4, 0\leq\rho\leq \cos\phi.

Section 12.7 Problem 25: Use spherical coordinates. Evaluate \iiint_E xe^{x^2+y^2+z^2} dV, where E is the portion of the unit ball x^2 +y^2 +z^2 \leq 1 that lies in the first octant.

Solution: The portion of the unit ball that lies in the first octant is described by 0\leq\theta\leq\pi/2, 0\leq\phi\leq\pi/2, 0\leq\rho\leq 1. Under the spherical coordinates, \iiint_E xe^{x^2+y^2+z^2} dV=\int_0^{\pi/2}\int_0^{\pi/2}\int_0^1 \rho\sin\phi\cos\theta e^{\rho^2}\rho^2\sin\phi d\rho d\phi d\theta = \int_0^{\pi/2} \cos\theta d\theta\int_0^{\pi/2} \sin^2\phi d\phi\int_0^1 \rho^3e^{\rho^2}d\rho = \frac{\pi}{8}.