# 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.