🌊 Unit 4 – Part B (Vector Calculus)

$$ \begin{aligned} \text{The normal vector to a surface } \phi \text{ is given by } \nabla \phi. \\ \textbf{Surface 1: } \phi_1 &= x^2 + y^2 + z^2 - 9 \\ \nabla \phi_1 &= \frac{\partial \phi_1}{\partial x}\vec{i} + \frac{\partial \phi_1}{\partial y}\vec{j} + \frac{\partial \phi_1}{\partial z}\vec{k} \\ &= 2x\vec{i} + 2y\vec{j} + 2z\vec{k} \\ \text{At } (2, -1, 2): \quad \vec{N}_1 &= 2(2)\vec{i} + 2(-1)\vec{j} + 2(2)\vec{k} = 4\vec{i} - 2\vec{j} + 4\vec{k} \\ \\ \textbf{Surface 2: } \phi_2 &= x^2 + y^2 - z - 3 \\ \nabla \phi_2 &= 2x\vec{i} + 2y\vec{j} - \vec{k} \\ \text{At } (2, -1, 2): \quad \vec{N}_2 &= 2(2)\vec{i} + 2(-1)\vec{j} - \vec{k} = 4\vec{i} - 2\vec{j} - \vec{k} \\ \\ \textbf{Angle } \theta: \quad \cos \theta &= \frac{\vec{N}_1 \cdot \vec{N}_2}{|\vec{N}_1| |\vec{N}_2|} \\ \vec{N}_1 \cdot \vec{N}_2 &= (4)(4) + (-2)(-2) + (4)(-1) = 16 + 4 - 4 = 16 \\ |\vec{N}_1| &= \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16+4+16} = \sqrt{36} = 6 \\ |\vec{N}_2| &= \sqrt{4^2 + (-2)^2 + (-1)^2} = \sqrt{16+4+1} = \sqrt{21} \\ \cos \theta &= \frac{16}{6\sqrt{21}} = \frac{8}{3\sqrt{21}} \\ \theta &= \cos^{-1}\left( \frac{8}{3\sqrt{21}} \right) \end{aligned} $$

$$ \begin{aligned} \textbf{1. Divergence (div } \vec{F}): \\ \text{div } \vec{F} &= \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xz^3) + \frac{\partial}{\partial y}(-2xyz) + \frac{\partial}{\partial z}(xz) \\ &= z^3 - 2xz + x \\ \text{At } (1, 2, 0): \quad &= (0)^3 - 2(1)(0) + 1 = 1 \\ \\ \textbf{2. Curl (curl } \vec{F}): \\ \text{curl } \vec{F} &= \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xz^3 & -2xyz & xz \end{vmatrix} \\ &= \vec{i} \left[ \frac{\partial}{\partial y}(xz) - \frac{\partial}{\partial z}(-2xyz) \right] - \vec{j} \left[ \frac{\partial}{\partial x}(xz) - \frac{\partial}{\partial z}(xz^3) \right] + \vec{k} \left[ \frac{\partial}{\partial x}(-2xyz) - \frac{\partial}{\partial y}(xz^3) \right] \\ &= \vec{i} [0 - (-2xy)] - \vec{j} [z - 3xz^2] + \vec{k} [-2yz - 0] \\ &= 2xy \vec{i} - (z - 3xz^2) \vec{j} - 2yz \vec{k} \\ \text{At } (1, 2, 0): \quad &= 2(1)(2) \vec{i} - (0 - 0) \vec{j} - 0 \vec{k} \\ &= 4\vec{i} \end{aligned} $$

$$ \begin{aligned} \text{Surfaces are orthogonal if } \nabla \phi \cdot \nabla \psi = 0. \\ \nabla \phi &= -1\vec{i} + 2y\vec{j} + 2z\vec{k} \\ \nabla \psi &= 4\vec{i} + \frac{1}{y^2+z^2}(2y)\vec{j} + \frac{1}{y^2+z^2}(2z)\vec{k} \\ &= 4\vec{i} + \frac{2y}{y^2+z^2}\vec{j} + \frac{2z}{y^2+z^2}\vec{k} \\ \text{Dot Product:} \\ \nabla \phi \cdot \nabla \psi &= (-1)(4) + (2y)\left(\frac{2y}{y^2+z^2}\right) + (2z)\left(\frac{2z}{y^2+z^2}\right) \\ &= -4 + \frac{4y^2}{y^2+z^2} + \frac{4z^2}{y^2+z^2} \\ &= -4 + \frac{4(y^2+z^2)}{y^2+z^2} \\ &= -4 + 4 = 0 \\ \text{Since the dot product is 0, the surfaces are orthogonal.} \end{aligned} $$

$$ \begin{aligned} \text{Irrotational means curl } \vec{F} = \vec{0}. \\ \nabla \times \vec{F} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x+2y+az & bx-3y-z & 4x+cy+2z \end{vmatrix} \\ &= \vec{i} \left[ \frac{\partial}{\partial y}(4x+cy+2z) - \frac{\partial}{\partial z}(bx-3y-z) \right] \\ &\quad - \vec{j} \left[ \frac{\partial}{\partial x}(4x+cy+2z) - \frac{\partial}{\partial z}(x+2y+az) \right] \\ &\quad + \vec{k} \left[ \frac{\partial}{\partial x}(bx-3y-z) - \frac{\partial}{\partial y}(x+2y+az) \right] \\ &= \vec{i} [c - (-1)] - \vec{j} [4 - a] + \vec{k} [b - 2] \\ &= (c+1)\vec{i} - (4-a)\vec{j} + (b-2)\vec{k} \\ \text{For vector to be zero, each component must be zero:} \\ c + 1 = 0 &\implies \mathbf{c = -1} \\ -(4 - a) = 0 &\implies \mathbf{a = 4} \\ b - 2 = 0 &\implies \mathbf{b = 2} \end{aligned} $$

$$ \begin{aligned} \text{Theorem: } \iint_S \vec{F} \cdot \vec{n} \, dS &= \iiint_V (\nabla \cdot \vec{F}) \, dV \\ \textbf{RHS (Volume Integral):} \\ \nabla \cdot \vec{F} &= \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3 \\ \iiint_V 3 \, dV &= 3 \int_0^a \int_0^a \int_0^a dx \, dy \, dz = 3(a \cdot a \cdot a) = 3a^3 \\ \\ \textbf{LHS (Surface Integral):} \text{ The cube has 6 faces.} \\ \text{1. Face } x=a \ (\vec{n}=\vec{i}): & \iint (x\vec{i}+y\vec{j}+z\vec{k}) \cdot \vec{i} \, dy dz = \int_0^a \int_0^a a \, dy dz = a(a^2) = a^3 \\ \text{2. Face } x=0 \ (\vec{n}=-\vec{i}): & \iint -x \, dy dz = \text{At } x=0, \text{integral is } 0. \\ \text{3. Face } y=a \ (\vec{n}=\vec{j}): & \iint (x\vec{i}+y\vec{j}+z\vec{k}) \cdot \vec{j} \, dx dz = \int_0^a \int_0^a a \, dx dz = a^3 \\ \text{4. Face } y=0 \ (\vec{n}=-\vec{j}): & \iint -y \, dx dz = 0. \\ \text{5. Face } z=a \ (\vec{n}=\vec{k}): & \iint (x\vec{i}+y\vec{j}+z\vec{k}) \cdot \vec{k} \, dx dy = \int_0^a \int_0^a a \, dx dy = a^3 \\ \text{6. Face } z=0 \ (\vec{n}=-\vec{k}): & \iint -z \, dx dy = 0. \\ \textbf{Total LHS} &= a^3 + 0 + a^3 + 0 + a^3 + 0 = 3a^3 \\ \text{LHS } &= \text{ RHS. Verified.} \end{aligned} $$

$$ \begin{aligned} \text{Using Gauss Divergence Theorem, convert to Volume Integral:} \\ I &= \iiint_V (\nabla \cdot \vec{F}) \, dV \\ \nabla \cdot \vec{F} &= \frac{\partial}{\partial x}(y-z+2) + \frac{\partial}{\partial y}(y+4) + \frac{\partial}{\partial z}(-xz) \\ &= 0 + 1 - x = 1 - x \\ I &= \int_0^2 \int_0^2 \int_0^2 (1-x) \, dx \, dy \, dz \\ \text{Separate integrals:} \\ I &= \left( \int_0^2 (1-x) dx \right) \times \left( \int_0^2 dy \right) \times \left( \int_0^2 dz \right) \\ \int_0^2 (1-x) dx &= \left[ x - \frac{x^2}{2} \right]_0^2 = \left( 2 - \frac{4}{2} \right) - 0 = 0 \\ I &= (0) \times (2) \times (2) = 0 \end{aligned} $$

$$ \begin{aligned} \textbf{RHS (Volume Integral):} \\ \nabla \cdot \vec{F} &= 2x + 2y + 2z \\ I &= \int_0^c \int_0^b \int_0^b (2x + 2y + 2z) \, dx \, dy \, dz \\ \text{Inner (x): } \int_0^b (2x+2y+2z) dx &= [x^2 + 2x(y+z)]_0^b = b^2 + 2b(y+z) \\ \text{Middle (y): } \int_0^b (b^2 + 2by + 2bz) dy &= [b^2y + by^2 + 2bzy]_0^b \\ &= b^3 + b^3 + 2b^2z = 2b^3 + 2b^2z \\ \text{Outer (z): } \int_0^c (2b^3 + 2b^2z) dz &= [2b^3z + b^2z^2]_0^c = 2b^3c + b^2c^2 \\ &= \mathbf{b^2c(2b + c)} \\ \\ \textbf{LHS (Surface Integral):} \\ \text{1. } x=b (\vec{n}=\vec{i}): \iint b^2 dy dz &= b^2(b)(c) = b^3c \\ \text{2. } x=0 (\vec{n}=-\vec{i}): \iint -0 dy dz &= 0 \\ \text{3. } y=b (\vec{n}=\vec{j}): \iint b^2 dx dz &= b^2(b)(c) = b^3c \\ \text{4. } y=0 (\vec{n}=-\vec{j}): \iint -0 dx dz &= 0 \\ \text{5. } z=c (\vec{n}=\vec{k}): \iint c^2 dx dy &= c^2(b)(b) = b^2c^2 \\ \text{6. } z=0 (\vec{n}=-\vec{k}): \iint -0 dx dy &= 0 \\ \textbf{Total} &= 2b^3c + b^2c^2 = \mathbf{b^2c(2b+c)} \end{aligned} $$

$$ \begin{aligned} \text{Theorem: } \oint_C \vec{F} \cdot d\vec{r} &= \iint_S (\nabla \times \vec{F}) \cdot \vec{n} \, dS \\ \textbf{RHS (Surface Integral):} \quad \vec{n}=\vec{k}, dS=dx dy \\ \nabla \times \vec{F} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \partial_x & \partial_y & \partial_z \\ x^2-y^2 & 2xy & 0 \end{vmatrix} \\ &= \vec{k} \left[ \frac{\partial}{\partial x}(2xy) - \frac{\partial}{\partial y}(x^2-y^2) \right] \\ &= \vec{k} [2y - (-2y)] = 4y\vec{k} \\ \iint_S (4y\vec{k}) \cdot \vec{k} \, dx dy &= \int_0^a \int_0^b 4y \, dy \, dx \\ &= \int_0^a \left[ 2y^2 \right]_0^b dx = \int_0^a 2b^2 dx = 2b^2[x]_0^a = \mathbf{2ab^2} \\ \\ \textbf{LHS (Line Integral):} \text{ 4 segments of rectangle.} \\ \text{1. } y=0 \text{ (from } x=0 \to a): \vec{F} \cdot d\vec{r} = (x^2-0)dx \implies \int_0^a x^2 dx = \frac{a^3}{3} \\ \text{2. } x=a \text{ (from } y=0 \to b): dx=0. \implies \int_0^b 2(a)y dy = a[y^2]_0^b = ab^2 \\ \text{3. } y=b \text{ (from } x=a \to 0): dy=0. \implies \int_a^0 (x^2-b^2)dx = \left[\frac{x^3}{3} - b^2x\right]_a^0 \\ = 0 - (\frac{a^3}{3} - ab^2) = -\frac{a^3}{3} + ab^2 \\ \text{4. } x=0 \text{ (from } y=b \to 0): dx=0. \text{ Also } x=0 \text{ makes term } 2xy=0. \implies \int_b^0 0 dy = 0 \\ \textbf{Total LHS} &= \frac{a^3}{3} + ab^2 - \frac{a^3}{3} + ab^2 + 0 = \mathbf{2ab^2} \end{aligned} $$

$$ \begin{aligned} \textbf{RHS (Surface):} \\ \nabla \times \vec{F} &= \vec{k} [\partial_x(xy) - \partial_y(x^2)] = \vec{k}[y - 0] = y\vec{k} \\ \iint y \, dx dy &= \int_0^a \int_0^a y \, dy dx = a \left[ \frac{y^2}{2} \right]_0^a = \frac{a^3}{2} \\ \\ \textbf{LHS (Line Integral):} \\ \text{1. } y=0 (x: 0 \to a): \int_0^a x^2 dx = \frac{a^3}{3} \\ \text{2. } x=a (y: 0 \to a): \int_0^a (a)y dy = a \frac{a^2}{2} = \frac{a^3}{2} \\ \text{3. } y=a (x: a \to 0): \int_a^0 x^2 dx = -\frac{a^3}{3} \\ \text{4. } x=0 (y: a \to 0): \text{Integrand } xy = 0 \implies 0 \\ \textbf{Total} &= \frac{a^3}{3} + \frac{a^3}{2} - \frac{a^3}{3} + 0 = \mathbf{\frac{a^3}{2}} \end{aligned} $$

$$ \begin{aligned} \text{Green's Theorem: } \oint P dx + Q dy &= \iint \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy \\ P = x^2-y^2, \quad Q = 4xy \\ \frac{\partial Q}{\partial x} = 4y, \quad \frac{\partial P}{\partial y} = -2y \\ \text{Integrand: } 4y - (-2y) = 6y \\ \textbf{RHS:} \int_0^a \int_0^b 6y \, dy dx &= \int_0^a \left[ 3y^2 \right]_0^b dx = \int_0^a 3b^2 dx = \mathbf{3ab^2} \\ \\ \textbf{LHS:} \\ \text{1. } y=0: \int_0^a x^2 dx = \frac{a^3}{3} \\ \text{2. } x=a: \int_0^b 4ay dy = 4a \frac{b^2}{2} = 2ab^2 \\ \text{3. } y=b: \int_a^0 (x^2-b^2) dx = \left[\frac{x^3}{3} - b^2x\right]_a^0 = -\frac{a^3}{3} + ab^2 \\ \text{4. } x=0: \int_b^0 0 dy = 0 \\ \textbf{Total} &= \frac{a^3}{3} + 2ab^2 - \frac{a^3}{3} + ab^2 = \mathbf{3ab^2} \end{aligned} $$

$$ \begin{aligned} \text{Intersections: } x^2=x \implies x=0, 1. \\ \textbf{RHS (Double Integral):} \\ Q=x^2, P=xy+y^2 \implies \partial_x Q = 2x, \partial_y P = x+2y \\ \text{Integrand: } 2x - (x+2y) = x-2y \\ \int_0^1 \int_{x^2}^x (x-2y) dy dx &= \int_0^1 \left[ xy - y^2 \right]_{x^2}^x dx \\ &= \int_0^1 [ (x^2-x^2) - (x^3-x^4) ] dx \\ &= \int_0^1 (x^4 - x^3) dx = \left[ \frac{x^5}{5} - \frac{x^4}{4} \right]_0^1 = \frac{1}{5} - \frac{1}{4} = \mathbf{-\frac{1}{20}} \\ \\ \textbf{LHS (Line Integral):} \\ \text{1. Along } y=x^2 (0 \to 1): dy=2xdx \\ \int_0^1 [ (x(x^2)+(x^2)^2) + x^2(2x) ] dx &= \int_0^1 (x^3+x^4+2x^3) dx = \int_0^1 (3x^3+x^4) dx \\ &= \frac{3}{4} + \frac{1}{5} = \frac{15+4}{20} = \frac{19}{20} \\ \text{2. Along } y=x (1 \to 0): dy=dx \\ \int_1^0 [ (x^2+x^2) + x^2 ] dx &= \int_1^0 3x^2 dx = [x^3]_1^0 = -1 \\ \textbf{Total} &= \frac{19}{20} - 1 = \mathbf{-\frac{1}{20}} \end{aligned} $$

$$ \begin{aligned} \text{Region: } y \text{ from } x^2 \text{ (bottom) to } \sqrt{x} \text{ (top)}. x \in [0,1]. \\ \textbf{RHS:} \quad P=3x^2-8y^2, Q=4y-6xy \\ \partial_x Q = -6y, \quad \partial_y P = -16y \\ \text{Integrand: } -6y - (-16y) = 10y \\ \int_0^1 \int_{x^2}^{\sqrt{x}} 10y \, dy dx &= \int_0^1 5[y^2]_{x^2}^{\sqrt{x}} dx = \int_0^1 5(x - x^4) dx \\ &= 5 \left[ \frac{x^2}{2} - \frac{x^5}{5} \right]_0^1 = 5 \left( \frac{1}{2} - \frac{1}{5} \right) = 5\left(\frac{3}{10}\right) = \mathbf{\frac{3}{2}} \\ \\ \textbf{LHS:} \\ \text{1. } y=x^2 (0 \to 1): dy=2xdx \\ I_1 = \int_0^1 [(3x^2-8x^4) + (4x^2-6x^3)2x] dx &= \int_0^1 (3x^2 - 8x^4 + 8x^3 - 12x^4) dx \\ &= \int_0^1 (3x^2 + 8x^3 - 20x^4) dx \\ &= 1 + \frac{8}{4} - \frac{20}{5} = 1 + 2 - 4 = -1 \\ \text{2. } y=\sqrt{x} \implies x=y^2 (y: 1 \to 0): dx=2ydy \\ I_2 = \int_1^0 [ (3y^4-8y^2)2y + (4y-6y^3) ] dy &= \int_1^0 (6y^5 - 16y^3 + 4y - 6y^3) dy \\ &= \int_1^0 (6y^5 - 22y^3 + 4y) dy \\ &= \left[ y^6 - \frac{22}{4}y^4 + 2y^2 \right]_1^0 \\ &= 0 - (1 - 5.5 + 2) = -(-2.5) = 2.5 = \frac{5}{2} \\ \textbf{Total} &= -1 + \frac{5}{2} = \mathbf{\frac{3}{2}} \end{aligned} $$