1. **Problem:** Find the domain, range, and graph of the function $f(x,y) = \sqrt{1 - x^2 - y^2}$. 2. **Domain:** The expression inside the square root must be non-negative: $$1 - x^2 - y^2 \geq 0 \implies x^2 + y^2 \leq 1$$ This describes the unit disk centered at the origin. 3. **Range:** Since the square root outputs non-negative values and the maximum inside is 1, the range is: $$0 \leq f(x,y) \leq 1$$ 4. **Graph:** The graph is the upper hemisphere of the unit sphere: $$z = \sqrt{1 - x^2 - y^2}$$ --- 1. **Problem:** Find the equation of the surface $z = \frac{1}{2}(x^2 + y^2)$ at point $P(-1, -1, 1)$. 2. This is a paraboloid. At $P$, verify: $$z = \frac{1}{2}((-1)^2 + (-1)^2) = \frac{1}{2}(1 + 1) = 1$$ 3. The surface is smooth and the point lies on it. --- 1. **Problem:** Compute partial derivatives $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \theta}$ for $$f(x,y) = \frac{y}{x^2 + y^2}, \quad x = r \cos \theta, \quad y = r \sin \theta$$ 2. Substitute: $$f(r,\theta) = \frac{r \sin \theta}{r^2} = \frac{\sin \theta}{r}$$ 3. Compute derivatives: $$\frac{\partial f}{\partial r} = -\frac{\sin \theta}{r^2}$$ $$\frac{\partial f}{\partial \theta} = \frac{\cos \theta}{r}$$ --- 1. **Problem:** Simplify and analyze $$f(x,y) = x^2 + y^2 - 2x + 2y$$ 2. Complete the square: $$x^2 - 2x + y^2 + 2y = (x^2 - 2x + 1) + (y^2 + 2y + 1) - 1 - 1 = (x-1)^2 + (y+1)^2 - 2$$ 3. This represents a circle centered at $(1,-1)$ with radius $\sqrt{2}$ shifted down by 2. --- 1. **Problem:** Evaluate the double integral $$\iint_D xy^2 \, dxdy$$ where $D$ is bounded by $\gamma_1: y=2x$ and $\gamma_2: y=x^2$. 2. Find intersection points: $$x^2 = 2x \implies x^2 - 2x = 0 \implies x(x-2)=0 \implies x=0,2$$ 3. Set up integral: $$\int_0^2 \int_{y=x^2}^{2x} x y^2 \, dy \, dx$$ 4. Integrate inner integral: $$\int_{x^2}^{2x} x y^2 \, dy = x \left[ \frac{y^3}{3} \right]_{x^2}^{2x} = x \left( \frac{(2x)^3}{3} - \frac{(x^2)^3}{3} \right) = x \left( \frac{8x^3}{3} - \frac{x^6}{3} \right) = \frac{8x^4 - x^7}{3}$$ 5. Integrate outer integral: $$\int_0^2 \frac{8x^4 - x^7}{3} \, dx = \frac{1}{3} \left[ \frac{8x^5}{5} - \frac{x^8}{8} \right]_0^2 = \frac{1}{3} \left( \frac{8 \cdot 32}{5} - \frac{256}{8} \right) = \frac{1}{3} \left( \frac{256}{5} - 32 \right) = \frac{1}{3} \left( \frac{256 - 160}{5} \right) = \frac{96}{15} = \frac{32}{5}$$ --- 1. **Problem:** Evaluate $$\iint_D \frac{x}{\sqrt{x^2 + y^2}} \, dxdy$$ where $D = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1, x \geq 0\}$. 2. Use polar coordinates: $$x = r \cos \theta, y = r \sin \theta, r \in [0,1], \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$$ 3. The integrand becomes: $$\frac{x}{\sqrt{x^2 + y^2}} = \frac{r \cos \theta}{r} = \cos \theta$$ 4. The Jacobian is $r$, so integral is: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^1 \cos \theta \cdot r \, dr \, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \left[ \frac{r^2}{2} \right]_0^1 d\theta = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \, d\theta$$ 5. Integrate: $$\frac{1}{2} [\sin \theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2} (1 - (-1)) = 1$$ --- 1. **Problem:** Solve the differential equation $$y' \tan x = y$$ 2. Rewrite: $$\frac{dy}{dx} = y \cot x$$ 3. Separate variables: $$\frac{dy}{y} = \cot x \, dx$$ 4. Integrate: $$\int \frac{1}{y} dy = \int \cot x \, dx \implies \ln |y| = \ln |\sin x| + C$$ 5. Exponentiate: $$y = A \sin x, \quad A = e^C$$ --- 1. **Problem:** Solve the third-order ODE $$y''' - 5y' + 4y = 0$$ with initial conditions $$y(0) = 5, \quad y'(0) = 8$$ 2. Characteristic equation: $$r^3 - 5r + 4 = 0$$ 3. Try roots: $r=1$: $$1 - 5 + 4 = 0$$ So $r=1$ is a root. 4. Factor: $$(r-1)(r^2 + r - 4) = 0$$ 5. Solve quadratic: $$r = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$$ 6. General solution: $$y = C_1 e^{x} + C_2 e^{\frac{-1 + \sqrt{17}}{2} x} + C_3 e^{\frac{-1 - \sqrt{17}}{2} x}$$ 7. Use initial conditions to solve for $C_1, C_2, C_3$ (requires $y''(0)$ or more info, not given, so only partial). --- **Final answers:** - Domain of $f(x,y) = \sqrt{1 - x^2 - y^2}$ is $x^2 + y^2 \leq 1$. - Range is $[0,1]$. - Partial derivatives $\frac{\partial f}{\partial r} = -\frac{\sin \theta}{r^2}$, $\frac{\partial f}{\partial \theta} = \frac{\cos \theta}{r}$. - Double integral over $D$ bounded by $y=2x$ and $y=x^2$ is $\frac{32}{5}$. - Double integral over semicircle $D$ is $1$. - Solution to $y' \tan x = y$ is $y = A \sin x$. - Characteristic roots for ODE are $1$, $\frac{-1 \pm \sqrt{17}}{2}$.