看题，北大高数期中试题

[swjtuer]

洗脚不要学习弃婴

你搞数学
如同弃婴搞普通物理

都是找打脸

你觉得这几道题难吗，就是考察一下知识点

[GreatCanada]

感觉美国高中学过AP的能及格

ChantGPT答案，一键生成，满分
Of course. Here are the detailed solutions to all the questions from the provided exam paper. ​1. (20 points) Calculate the integrals: ​(1) I = \iint_D \sqrt{|y - x²|} \,dx\,dy, where D is the rectangular region [-1, 1] \times [0, 2]. ​Solution: The domain of integration is D = \{(x, y) \mid -1 \le x \le 1, 0 \le y \le 2\}. The integrand contains an absolute value term |y - x²|, so we need to split the integral based on the sign of y - x². The curve y = x² divides the rectangle D into two regions: ​D_1 = \{(x, y) \in D \mid y \ge x^2\} ​D_2 = \{(x, y) \in D \mid y < x^2\} ​The integral becomes: I = \iint{D_1} \sqrt{y - x²} \,dx\,dy + \iint{D_2} \sqrt{x² - y} \,dx\,dy ​It's easier to integrate with respect to y first. I = \int{-1}^{1} \left[ \int{0}^{2} \sqrt{|y - x²|} \,dy \right] dx ​For a fixed x \in [-1, 1], the value of x² is in [0, 1]. We split the inner integral at y = x²: \int{0}^{2} \sqrt{|y - x²|} \,dy = \int{0}^{x²} \sqrt{x² - y} \,dy + \int{x²}^{2} \sqrt{y - x²} \,dy ​For the first part: Let u = x² - y, so du = -dy. \int{0}^{x²} \sqrt{x² - y} \,dy = \int{x²}^{0} \sqrt{u} (-du) = \int{0}^{x²} u^{1/2} \,du = \left[ \frac{2}{3}u^{3/2} \right]{0}^{x²} = \frac{2}{3}(x²)^{3/2} = \frac{2}{3}x³ (since x \in [-1,1], this should be |x|³). Let's be careful. The result is \frac{2}{3}|x³|. ​For the second part: Let v = y - x², so dv = dy. \int{x²}^{2} \sqrt{y - x²} \,dy = \int{0}^{2-x²} \sqrt{v} \,dv = \left[ \frac{2}{3}v^{3/2} \right]{0}^{2-x²} = \frac{2}{3}(2 - x²)^{3/2} ​Now, we integrate the sum with respect to x from -1 to 1: I = \int{-1}^{1} \left( \frac{2}{3}|x|³ + \frac{2}{3}(2 - x²)^{3/2} \right) dx Since the integrand is an even function, we can integrate from 0 to 1 and multiply by 2: I = 2 \cdot \frac{2}{3} \int{0}^{1} \left( x³ + (2 - x²)^{3/2} \right) dx I = \frac{4}{3} \left[ \int{0}^{1} x³ \,dx + \int{0}^{1} (2 - x²)^{3/2} \,dx \right] ​The first integral is \int{0}^{1} x³ \,dx = \left[ \frac{x⁴}{4} \right]0¹ = \frac{1}{4}. ​For the second integral, let x = \sqrt{2}\sin\theta. Then dx = \sqrt{2}\cos\theta \,d\theta. When x=0, \theta=0. When x=1, \sin\theta = 1/\sqrt{2}, so \theta = \pi/4. \int{0}^{1} (2 - x²)^{3/2} \,dx = \int{0}^{\pi/4} (2 - 2\sin^2\theta)^{3/2} (\sqrt{2}\cos\theta) \,d\theta = \int{0}^{\pi/4} (2\cos^2\theta)^{3/2} (\sqrt{2}\cos\theta) \,d\theta = \int{0}^{\pi/4} (2\sqrt{2}\cos^3\theta) (\sqrt{2}\cos\theta) \,d\theta = 4 \int{0}^{\pi/4} \cos^4\theta \,d\theta = 4 \int{0}^{\pi/4} \left(\frac{1 + \cos(2\theta)}{2}\right)² \,d\theta = \int{0}^{\pi/4} (1 + 2\cos(2\theta) + \cos²(2\theta)) \,d\theta = \int{0}^{\pi/4} \left(1 + 2\cos(2\theta) + \frac{1 + \cos(4\theta)}{2}\right) \,d\theta = \int{0}^{\pi/4} \left(\frac{3}{2} + 2\cos(2\theta) + \frac{1}{2}\cos(4\theta)\right) \,d\theta = \left[ \frac{3}{2}\theta + \sin(2\theta) + \frac{1}{8}\sin(4\theta) \right]0^{\pi/4} = \left( \frac{3}{2}\frac{\pi}{4} + \sin(\frac{\pi}{2}) + \frac{1}{8}\sin(\pi) \right) - 0 = \frac{3\pi}{8} + 1 ​Finally, I = \frac{4}{3} \left( \frac{1}{4} + \frac{3\pi}{8} + 1 \right) = \frac{4}{3} \left( \frac{5}{4} + \frac{3\pi}{8} \right) = \frac{5}{3} + \frac{\pi}{2}. ​(2) I = \int{-1}^{1} dx \int{0}^{\sqrt{1-x²}} dy \int{1}^{1+\sqrt{1-x²-y²}} \frac{z}{x²+y²+z²} dz ​Solution: The region of integration for (x,y) is the upper semi-disk of radius 1: D{xy} = \{(x,y) \mid x²+y² \le 1, y \ge 0\}. The limits for z are from the plane z=1 to the sphere z = 1 + \sqrt{1-x²-y²}, which can be written as x²+y²+(z-1)² = 1. This is a sphere of radius 1 centered at (0,0,1). The integration volume is the part of the upper hemisphere of this sphere that lies above D{xy}. ​We use spherical coordinates centered at the origin: x = r\sin\phi\cos\theta, y = r\sin\phi\sin\theta, z = r\cos\phi. The Jacobian is r^2\sin\phi. The condition y \ge 0 implies \sin\theta \ge 0, so 0 \le \theta \le \pi. The plane z=1 is r\cos\phi = 1, or r = \sec\phi. The sphere x²+y²+z²-2z=0 is r² - 2r\cos\phi = 0, or r=2\cos\phi. The region is bounded by these two surfaces, so r ranges from \sec\phi to 2\cos\phi. For this range to be valid, we need \sec\phi \le 2\cos\phi \implies 1 \le 2\cos^2\phi \implies \cos\phi \ge 1/\sqrt{2}. Since we are in the upper-half space, 0 \le \phi \le \pi/2, so this gives 0 \le \phi \le \pi/4. ​The integrand is \frac{z}{x²+y²+z²} = \frac{r\cos\phi}{r²} = \frac{\cos\phi}{r}. The integral becomes: I = \int_0^\pi \int_0^{\pi/4} \int{\sec\phi}^{2\cos\phi} \left(\frac{\cos\phi}{r}\right) r^2\sin\phi \,dr\,d\phi\,d\theta I = \int_0^\pi d\theta \int_0^{\pi/4} \cos\phi\sin\phi \left[ \int{\sec\phi}^{2\cos\phi} r \,dr \right] d\phi The inner integral is \left[ \frac{r²}{2} \right]{\sec\phi}^{2\cos\phi} = \frac{1}{2}(4\cos^2\phi - \sec^2\phi). So, I = \pi \int_0^{\pi/4} \cos\phi\sin\phi \cdot \frac{1}{2}(4\cos^2\phi - \sec^2\phi) \,d\phi I = \frac{\pi}{2} \int_0^{\pi/4} (4\cos^{3\phi\sin\phi} - \tan\phi) \,d\phi Let's integrate term by term: \int 4\cos^{3\phi\sin\phi} \,d\phi: Let u=\cos\phi, du=-\sin\phi\,d\phi. \int -4u³ du = -u⁴ = -\cos^4\phi. \int -\tan\phi \,d\phi = \ln|\cos\phi|. So, the integral with respect to \phi is: \frac{\pi}{2} \left[ -\cos^4\phi + \ln(\cos\phi) \right]0^{\pi/4} At \phi=\pi/4: -\cos⁴(\pi/4) + \ln(\cos(\pi/4)) = -(1/\sqrt{2})⁴ + \ln(1/\sqrt{2}) = -1/4 - \frac{1}{2}\ln 2. At \phi=0: -\cos⁴(0) + \ln(\cos(0)) = -1 + 0 = -1. The result is \frac{\pi}{2} \left[ (-1/4 - \frac{1}{2}\ln 2) - (-1) \right] = \frac{\pi}{2} \left( \frac{3}{4} - \frac{1}{2}\ln 2 \right) = \pi \left( \frac{3}{8} - \frac{1}{4}\ln 2 \right). ​2. (10 points) Calculate the line integral I = \int{\Gamma} \left(\frac{1}{2} - \frac{y}{(x-\frac{1}{2})² + y²}\right)dx + \left(\frac{x-\frac{1}{2}}{(x-\frac{1}{2})² + y²}\right)dy, where \Gamma is the curve segment from (0, -1) along the parabola y² = 1-x to (0, 1). ​Solution: Let the vector field be \vec{F} = (P, Q), where P(x, y) = \frac{1}{2} - \frac{y}{(x-\frac{1}{2})² + y²} Q(x, y) = \frac{x-\frac{1}{2}}{(x-\frac{1}{2})² + y²} ​Let's check if the field is conservative by testing if \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}. \frac{\partial Q}{\partial x} = \frac{1 \cdot ((x-\frac{1}{2})² + y²) - (x-\frac{1}{2}) \cdot 2(x-\frac{1}{2})}{((x-\frac{1}{2})² + y²)²} = \frac{y² - (x-\frac{1}{2})²}{((x-\frac{1}{2})² + y²)²} \frac{\partial P}{\partial y} = - \frac{1 \cdot ((x-\frac{1}{2})² + y²) - y \cdot 2y}{((x-\frac{1}{2})² + y²)²} = - \frac{(x-\frac{1}{2})² - y²}{((x-\frac{1}{2})² + y²)²} = \frac{y² - (x-\frac{1}{2})²}{((x-\frac{1}{2})² + y²)²} ​Since \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}, the vector field is conservative in any simply connected region. The field is singular only at the point (1/2, 0). The path \Gamma is the parabola x=1-y² from (0,-1) to (0,1). The point (1/2, 0) is not on this path, as x=1/2 would imply y²=1/2. A simply connected region can be drawn around the path that excludes the singularity. ​Therefore, the integral is path-independent. We can find a potential function f(x, y) such that \nabla f = \vec{F}. \frac{\partial f}{\partial y} = Q(x,y) \implies f(x,y) = \int \frac{x-\frac{1}{2}}{(x-\frac{1}{2})² + y²} dy = \arctan\left(\frac{y}{x-\frac{1}{2}}\right) + g(x) \frac{\partial f}{\partial x} = \frac{-\frac{y}{(x-\frac{1}{2})²}}{1+(\frac{y}{x-\frac{1}{2}})²} + g'(x) = -\frac{y}{(x-\frac{1}{2})²+y²} + g'(x) We need this to equal P(x,y) = \frac{1}{2} - \frac{y}{(x-\frac{1}{2})² + y²}. Thus, g'(x) = 1/2, which gives g(x) = x/2. The potential function is f(x, y) = \frac{x}{2} + \arctan\left(\frac{y}{x-\frac{1}{2}}\right). ​The value of the integral is the difference of the potential function at the endpoints: I = f(0, 1) - f(0, -1) f(0, 1) = \frac{0}{2} + \arctan\left(\frac{1}{0-\frac{1}{2}}\right) = \arctan(-2) = -\arctan(2) f(0, -1) = \frac{0}{2} + \arctan\left(\frac{-1}{0-\frac{1}{2}}\right) = \arctan(2) I = -\arctan(2) - \arctan(2) = -2\arctan(2). ​3. (10 points) Calculate the surface integral I = \iint{\Sigma} 2xy \,dy\,dz - y² \,dz\,dx - x² \,dx\,dy, where \Sigma is the part of the paraboloid x² + y² = z between z=0 and z=1, with the outward side. ​Solution: This is a flux integral \iint\Sigma \vec{F} \cdot d\vec{S}, where \vec{F} = (2xy, -y², -x²). The surface \Sigma is not closed. "Outward side" means the normal vector points away from the z-axis. For the bowl-shaped surface z=x²+y², this corresponds to a normal vector with a negative z-component (pointing "down and out"). ​We can use the Divergence Theorem by closing the surface. Let \Sigma_1 be the disk x²+y² \le 1 at z=1. Let S = \Sigma \cup \Sigma_1. Let V be the volume enclosed by S. The outward normal for V points up on \Sigma_1 and "down and out" on \Sigma. The specified orientation for \Sigma is consistent with the outward normal from V. ​The Divergence Theorem states \oiint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F}) \,dV. Let's compute the divergence of \vec{F}: \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2xy) + \frac{\partial}{\partial y}(-y²) + \frac{\partial}{\partial z}(-x²) = 2y - 2y + 0 = 0. ​Since the divergence is zero, the triple integral is zero: \oiint_S \vec{F} \cdot d\vec{S} = 0. This means \iint{\Sigma} \vec{F} \cdot d\vec{S} + \iint{\Sigma_1} \vec{F} \cdot d\vec{S} = 0. So, the desired integral is I = \iint{\Sigma} \vec{F} \cdot d\vec{S} = - \iint{\Sigma_1} \vec{F} \cdot d\vec{S}. ​On the surface \Sigma_1, we have z=1 and the outward normal vector is \vec{n} = (0, 0, 1). So d\vec{S} = \vec{k} \,dx\,dy. \iint{\Sigma_1} \vec{F} \cdot d\vec{S} = \iint{x²+y² \le 1} \vec{F}(x,y,1) \cdot (0,0,1) \,dx\,dy = \iint{x²+y² \le 1} -x² \,dx\,dy. ​We evaluate this integral using polar coordinates: x = r\cos\theta, y=r\sin\theta, dx\,dy = r\,dr\,d\theta. \iint{x²+y² \le 1} -x² \,dx\,dy = \int_0^{2\pi} \int_0¹ -(r\cos\theta)² \cdot r \,dr\,d\theta = -\int_0^{2\pi} \int_0¹ r³ \cos^2\theta \,dr\,d\theta = -\left(\int_0¹ r³ \,dr\right) \left(\int_0^{2\pi} \cos^2\theta \,d\theta\right) = -\left[\frac{r⁴}{4}\right]0¹ \cdot \int_0^{2\pi} \frac{1+\cos(2\theta)}{2} \,d\theta = -\frac{1}{4} \cdot \left[\frac{\theta}{2} + \frac{\sin(2\theta)}{4}\right]0^{2\pi} = -\frac{1}{4} \cdot \pi = -\frac{\pi}{4}. ​Therefore, I = -(-\pi/4) = \pi/4. ​4. (10 points) Find the area of the region enclosed by the curve x³ + y³ = xy. ​Solution: This curve is a Folium of Descartes. It forms a loop in the first quadrant. We can calculate the area using Green's Theorem, A = \frac{1}{2} \oint_C x\,dy - y\,dx. ​We use the parameterization y=tx. Substituting into the equation: x³ + (tx)³ = x(tx) \implies x³(1+t³) = tx². Assuming x \ne 0, we get x = \frac{t}{1+t³} and y = tx = \frac{t²}{1+t³}. The loop is traced as t goes from 0 to \infty. ​Now we compute dx and dy: dx = \frac{1(1+t³) - t(3t²)}{(1+t³)²} dt = \frac{1-2t³}{(1+t³)²} dt dy = \frac{2t(1+t³) - t²(3t²)}{(1+t³)²} dt = \frac{2t-t⁴}{(1+t³)²} dt ​Next, we compute x\,dy - y\,dx: x\,dy - y\,dx = \frac{t}{1+t³} \frac{2t-t⁴}{(1+t³)²} dt - \frac{t²}{1+t³} \frac{1-2t³}{(1+t³)²} dt = \frac{t(2t-t⁴) - t²(1-2t³)}{(1+t³)³} dt = \frac{2t² - t⁵ - t² + 2t⁵}{(1+t³)³} dt = \frac{t² + t⁵}{(1+t³)³} dt = \frac{t²(1+t³)}{(1+t³)³} dt = \frac{t²}{(1+t³)²} dt ​The area is: A = \frac{1}{2} \int_0^\infty \frac{t²}{(1+t³)²} dt Let u = 1+t³, so du = 3t² dt, and t² dt = du/3. When t=0, u=1. As t \to \infty, u \to \infty. A = \frac{1}{2} \int_1^\infty \frac{1}{u²} \frac{du}{3} = \frac{1}{6} \int_1^\infty u^{-2} du = \frac{1}{6} \left[ -\frac{1}{u} \right]1^\infty = \frac{1}{6} (0 - (-1)) = \frac{1}{6}. ​5. (10 points) Find the volume of the solid enclosed by the rotating paraboloid z = 6 - x² - y² and the cone z = \sqrt{x² + y²}. ​Solution: We use cylindrical coordinates, where x=r\cos\theta, y=r\sin\theta, z=z. The equations of the surfaces become: Paraboloid: z = 6 - r² Cone: z = r ​The solid is bounded below by the cone and above by the paraboloid. To find the region of integration for r, we find the intersection of the two surfaces: r = 6 - r² \implies r² + r - 6 = 0 \implies (r+3)(r-2) = 0. Since radius r must be non-negative, we have r=2. The intersection is a circle of radius 2 at height z=2. ​The volume is given by the triple integral V = \iiint_E dV. In cylindrical coordinates, dV = r \,dz\,dr\,d\theta. The limits are: ​0 \le \theta \le 2\pi (by symmetry) ​0 \le r \le 2 ​r \le z \le 6-r² ​V = \int_0^{2\pi} \int_0² \int_r^{6-r²} r \,dz\,dr\,d\theta V = \int_0^{2\pi} \int_0² r [z]r^{6-r²} \,dr\,d\theta = \int_0^{2\pi} \int_0² r(6-r²-r) \,dr\,d\theta = \int_0^{2\pi} \int_0² (6r - r³ - r²) \,dr\,d\theta The inner integral is: \left[ 3r² - \frac{r⁴}{4} - \frac{r³}{3} \right]0² = (3(4) - \frac{16}{4} - \frac{8}{3}) - 0 = 12 - 4 - \frac{8}{3} = 8 - \frac{8}{3} = \frac{16}{3}. Now the outer integral: V = \int_0^{2\pi} \frac{16}{3} d\theta = \frac{16}{3} [\theta]0^{2\pi} = \frac{16}{3} (2\pi) = \frac{32\pi}{3}. ​6. (20 points) Solve the following differential equations: ​(1) y' - y = xy⁵ ​Solution: This is a Bernoulli equation of the form y' + P(x)y = Q(x)yⁿ with n=5. Note that y=0 is a trivial solution. For y \ne 0, we divide by y⁵: y^{-5}y' - y^{-4} = x. Let v = y^{-4}. Then v' = -4y^{-5}y'. So y^{-5}y' = -v'/4. Substituting this into the equation gives: -\frac{v'}{4} - v = x \implies v' + 4v = -4x. This is a first-order linear ODE. The integrating factor is I(x) = e^{\int 4 dx} = e^{4x}. Multiplying the equation by e^{4x}: e^{4x}v' + 4e^{4x}v = -4xe^{4x} \implies (e^{4x}v)' = -4xe^{4x}. Integrate both sides: e^{4x}v = \int -4xe^{4x} dx. We use integration by parts (\int u\,dv = uv - \int v\,du) with u=-4x, dv=e^{4x}dx. Then du=-4dx, v=\frac{1}{4}e^{4x}. \int -4xe^{4x} dx = -4x \cdot \frac{1}{4}e^{4x} - \int \frac{1}{4}e^{4x}(-4dx) = -xe^{4x} + \int e^{4x}dx = -xe^{4x} + \frac{1}{4}e^{4x} + C. So, e^{4x}v = -xe^{4x} + \frac{1}{4}e^{4x} + C. v = -x + \frac{1}{4} + Ce^{-4x}. Substituting back v=y^{-4}: y^{-4} = \frac{1}{4} - x + Ce^{-4x}. The solutions are y⁴ = \frac{1}{\frac{1}{4} - x + Ce^{-4x}} and the trivial solution y=0. ​(2) y'' + y' = 4\sin x + x\cos(2x) ​Solution: This is a second-order linear non-homogeneous ODE. ​Solve the homogeneous equation: y'' + y' = 0. The characteristic equation is r²+r=0 \implies r(r+1)=0, with roots r_1=0, r_2=-1. The complementary solution is y_c = C_1e^{0x} + C_2e^{-x} = C_1 + C_2e^{-x}. ​Find a particular solution y_p: We use the method of undetermined coefficients. We find a particular solution for each part of the right-hand side and add them up (y_p = y{p1} + y{p2}). ​For 4\sin x: Guess y{p1} = A\sin x + B\cos x. y'{p1} = A\cos x - B\sin x y''{p1} = -A\sin x - B\cos x Substituting into y''+y'=4\sin x: (-A\sin x - B\cos x) + (A\cos x - B\sin x) = 4\sin x (-A-B)\sin x + (A-B)\cos x = 4\sin x. Comparing coefficients: -A-B=4 and A-B=0 \implies A=B. -2A=4 \implies A=-2. So A=B=-2. y{p1} = -2\sin x - 2\cos x. ​For x\cos(2x): Guess y{p2} = (Cx+D)\cos(2x) + (Ex+F)\sin(2x). y'{p2} = (C+2Ex+2F)\cos(2x) + (-2Cx-2D+E)\sin(2x). y''{p2} = (-4Cx-4D+4E)\cos(2x) + (-4Ex-4F-4C)\sin(2x). y''{p2} + y'{p2} = [(-4C+2E)x + (C-4D+2F+4E)]\cos(2x) + [(-2C-4E)x + (-4C-2D-4F+E)]\sin(2x) = x\cos(2x). Comparing coefficients of terms: ​x\cos(2x): -4C+2E=1 ​x\sin(2x): -2C-4E=0 \implies C=-2E ​\cos(2x): C-4D+2F+4E=0 ​\sin(2x): -4C-2D-4F+E=0 From the first two equations: -4(-2E)+2E=1 \implies 10E=1 \implies E=1/10. Then C = -2/10 = -1/5. Substitute C and E into the last two equations: ​-1/5 - 4D + 2F + 4/10 = 0 \implies -4D+2F = -1/5 \implies 4D-2F=1/5. ​-4(-1/5) - 2D - 4F + 1/10 = 0 \implies 4/5+1/10=2D+4F \implies 9/10=2D+4F. Solving the system: 8D-4F = 2/5 (multiply first eq by 2) 2D+4F = 9/10 Adding them gives 10D = 2/5 + 9/10 = 13/10 \implies D=13/100. 2F = 4D - 1/5 = 4(13/100) - 20/100 = 32/100 \implies F = 16/100 = 4/25. y{p2} = (-\frac{1}{5}x + \frac{13}{100})\cos(2x) + (\frac{1}{10}x + \frac{4}{25})\sin(2x). ​General Solution: y = y_c + y{p1} + y{p2}. y = C_1 + C_2e^{-x} - 2\sin x - 2\cos x + (-\frac{1}{5}x + \frac{13}{100})\cos(2x) + (\frac{1}{10}x + \frac{4}{25})\sin(2x). ​7. (10 points) Let \Gamma be the counter-clockwise circle (x-1)² + (y-1)² = 1, and f(x) be a positive continuous function. Prove that: \oint\Gamma xf(y) \,dy - \frac{y}{f(x)} \,dx \ge 2\pi. ​Solution: Let the integral be I. We use Green's Theorem: \oint\Gamma P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy. Here, P = -y/f(x) and Q = xf(y). The region D is the disk (x-1)² + (y-1)² \le 1. \frac{\partial Q}{\partial x} = f(y) \frac{\partial P}{\partial y} = -1/f(x) So, the integral becomes: I = \iint_D \left( f(y) - \left(-\frac{1}{f(x)}\right) \right) dx\,dy = \iint_D \left( f(y) + \frac{1}{f(x)} \right) dx\,dy. ​The disk D is symmetric with respect to the line y=x. Because of this symmetry, we can swap the variables x and y in the integration over D: \iint_D f(y) \,dx\,dy = \iint_D f(x) \,dx\,dy. So we can rewrite the integral as: I = \iint_D f(x) \,dx\,dy + \iint_D \frac{1}{f(x)} \,dx\,dy = \iint_D \left( f(x) + \frac{1}{f(x)} \right) dx\,dy. ​The problem states that f(x) is a positive function, so f(x) > 0. By the AM-GM inequality, for any positive number a, a + \frac{1}{a} \ge 2\sqrt{a \cdot \frac{1}{a}} = 2. Thus, the integrand f(x) + \frac{1}{f(x)} \ge 2 for all x in the domain. ​Therefore, I = \iint_D \left( f(x) + \frac{1}{f(x)} \right) dx\,dy \ge \iint_D 2 \,dx\,dy. The integral \iint_D 2 \,dx\,dy is twice the area of the disk D. The disk D has radius r=1, so its area is \pi r² = \pi. I \ge 2 \cdot \text{Area}(D) = 2\pi. This completes the proof. ​8. (10 points) Assume f(t) is a continuous function. For t>0, let F(t) = \frac{\iiint{x²+y²+z² \le t²} f(x²+y²+z²) \,dV}{\iint{x²+y²+z² \le t²} (x²+y²)f(x²+y²+z²) \,d\sigma}. Prove that F(t) is a strictly decreasing function. ​Solution: There appears to be a typo in the problem statement. The denominator is a double integral over a solid ball, with a differential d\sigma. This is non-standard. The most likely interpretations are either: a) The integral is a surface integral over the boundary sphere, ∬{x²+y²+z²=t²} ... dσ. b) The integral is a volume integral over the ball, ∭{x²+y²+z² \le t²} ... dV. ​Interpretation (a) leads to a proof that requires differentiability of f, which is not given. Interpretation (b) leads to a clean proof without this requirement. We will proceed assuming the typo and that the denominator is a volume integral. We also assume f(t)>0, which is standard for such problems to ensure the denominator is non-zero and for the proof to hold. ​Let's evaluate the numerator N(t) and denominator D(t) using spherical coordinates (dV = r^2\sin\phi\,dr\,d\phi\,d\theta). N(t) = \iiint{r \le t} f(r²) \,dV = \int_0^t \int_0^{2\pi} \int_0^\pi f(r²) r^2\sin\phi \,d\phi\,d\theta\,dr N(t) = \left(\int_0^t r^2f(r²)\,dr\right) \left(\int_0^{2\pi}d\theta\right) \left(\int_0^\pi \sin\phi\,d\phi\right) = 4\pi \int_0^t r^2f(r²)\,dr. ​D(t) = \iiint{r \le t} (x²+y²)f(r²) \,dV = \iiint_{r \le t} (r^2\sin2\phi)f(r²) \,dV D(t) = \int_0^t \int_0^{2\pi} \int_0^\pi (r^2\sin2\phi)f(r²) r^2\sin\phi \,d\phi\,d\theta\,dr D(t) = \left(\int_0^t r^4f(r²)\,dr\right) \left(\int_0^{2\pi}d\theta\right) \left(\int_0^\pi \sin^3\phi\,d\phi\right). The integral \int_0^\pi \sin^3\phi\,d\phi = 4/3. So the angular part is 2\pi \cdot (4/3) = 8\pi/3. D(t) = \frac{8\pi}{3} \int_0^t r^4f(r²)\,dr. ​So, F(t) = \frac{4\pi \int_0^t r^2f(r²)\,dr}{\frac{8\pi}{3} \int_0^t r^4f(r²)\,dr} = \frac{3 \int_0^t r^2f(r²)\,dr}{2 \int_0^t r^4f(r²)\,dr}. Let g(r) = r^2f(r²). Since f>0, g(r)>0 for r>0. F(t) = \frac{3}{2} \frac{\int_0^t g(r)\,dr}{\int_0^t r^2g(r)\,dr}. To prove F(t) is strictly decreasing, we need to show F'(t) < 0. Let's examine the derivative of the ratio o