# Recitation 13

Section 12.7 Problem 23: Evaluate $\iiint_E (x^2 + y^2) dV$, where $E$ lies between the spheres $x^2 +y^2 +z^2 =4$ and $x^2 +y^2 +z^2 =9$.

Solution: By symmetry, $\iint_E (x^2+y^2) dV = \iint_E (y^2+z^2) dV = \iint_E (z^2+x^2) dV$. Therefore $\iint_E (x^2+y^2) dV = \frac{2}{3}\iint_E (x^2+y^2+z^2) dV$. Turn this into spherical coordinates, we get $\frac{2}{3}\int_0^{2\pi}\int_0^\pi\int_2^3\rho^2\rho^2\sin\phi d\rho d\phi d\theta = \frac{1688}{15}\pi$.

Section 12.7 Problem 27 (a): Find the volume of the solid that lies above the cone $\phi = \pi / 3$ and below the sphere $\rho = 4 \cos \phi$.

Comment: Under spherical coordinates, the volume is equal to $\int_0^{2\pi}\int_0^{\pi/3}\int_0^{4\cos\phi}\rho^2\sin\phi d\rho d\phi d\theta$.

Section 13.2 Problem 12: Evaluate the line integral, where $C$ is the given curve. $\int_C(x^2+y^2+z^2)ds$, $C: x=t, y=\cos 2t, z=\sin 2t, 0\leq t\leq 2\pi$.

Comment: Under the parametrization of the curve, we obtain $\int_C(x^2+y^2+z^2)ds = \int_0^{2\pi}(t^2+1)|r'(t)|dt$. Because $r'(t)=(1, -2\sin 2t, 2\cos 2t)$, $|r'(t)| = \sqrt{5}$. Therefore $\int_0^{2\pi}(t^2+1)|r'(t)|dt = \int_0^{2\pi}\sqrt{5}(t^2+1)dt = \sqrt{5} (2\pi + 8\pi^3/3)$.

Section 13.2 Problem 16: Evaluate the line integral, where $C$ is the given curve. $\int_C (y+z)dx+(x+z)dy+(x+y)dz$, $C$ consists of line segments from $(0, 0, 0)$ to $(1, 0, 1)$ and from $(1, 0, 1)$ to $(0, 1, 2)$.

Solution: Since $(y+z, x+z, x+y)$ is the gradient of $f(x,y,z)=xy+yz+zx$. The line integral is equal to $f(0,1,2)-f(0,0,0)=2$.

Section 13.2 Problem 18: The figure shows a vector field $F$ and two curves $C_1$ and $C_2$. Are the line integrals of $F$ over $C_1$ and $C_2$ positive, negative, or zero? Explain.

Comment: Please refer to the textbook for the figure.

Section 13.2 Problem 22: Evaluate the line integral $\int_C F \cdot dr$, where $C$ is given by the vector function $r(t)$. $F(x,y,z)=xi+yj+xyk$, $r(t)=\cos ti+\sin tj+tk$, $0\leq t\leq \pi$.

Solution: Under the parametrization of the curve, we obtain $\int_C F \cdot dr=\int_0^\pi (\cos ti + \sin t j + \cos t\sin t k)\cdot (-\sin ti+\cos tj+k)dt = 0$.

Section 13.2 Problem 38: Find the work done by the force field $F(x, y) = x^2 i + ye^x j$ on a particle that moves along the parabola $x = y^2+1$ from $(1, 0)$ to $(2, 1)$.

Solution: Parametrize the curve by $x=t^2+1, y=t$ with $0\leq t\leq 1$. The work is equal to $\int_0^1 ((t^2+1)^2, te^{t^2+1})\cdot (2t, 1)dt = \frac{7}{3}+\frac{1}{2}(e-1)e$.