# 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, $\iiint_E xe^{x^2+y^2+z^2} dV=\int_0^{\pi/2}\int_0^{\pi/2}\int_0^1 \rho\sin\phi\cos\theta e^{\rho^2}\rho^2\sin\phi d\rho d\phi d\theta = \int_0^{\pi/2} \cos\theta d\theta\int_0^{\pi/2} \sin^2\phi d\phi\int_0^1 \rho^3e^{\rho^2}d\rho = \frac{\pi}{8}$.