Multivariable calculus

1. Definitions:

• \(f: \RR^2 \to \RR, \quad (x,y) \mapsto z = f(x,y)\)

• Level curves are given by \(f(x, y) = c\);

• \(\nabla f: \RR^2 \to \RR^2, \quad \nabla f(\vector p) = \thevector{f_x(\vector p), f_y(\vector p)}\)

• \(\gamma(t): \RR \to \RR^2, \quad \gamma(t) = \thevector{x(t), y(t)}, \quad \gamma_t(t) = \thevector{x_t(t), y_t(t)}\)

• \(g(t) = f \circ \gamma:\RR \to \RR, \quad (f \circ \gamma)(t) = f(x(t), y(t)), \quad g_t(t) = f_t(\gamma(t))\cdot \gamma_t(t)\)

• A vector \(v\) is perpendicular to a surface at a point \(p\) iff \(v\) is perpendicular to the tangent vector of every curve passing through \(p\).

• Proof of actual statement

  • Let \(\vector p = \thevector{x_0, y_0}\) be a point on the level surface, so \(f(\vector p) = c\) for some constant.

  • Let \(\gamma(t) = \thevector{x(t), y(t)}\) be a curve on the level surface, so \(\gamma(t_0) = \vector p\) for some \(t_0\). Let \(\gamma'(t)\) be its tangent vector.

  • Let \(g(t) = f(x(t), y(t)) = (f \circ \gamma)(t)\), and note that for some \(\vector x\) on the level surface, and so we have \(g(t_0) = f(\vector x) = c\) and thus \(\dd{g}{t}(t) = 0\).

  • By the chain rule, compute \[\begin{align*} \dd{g}{t}(t) &= \left(\dd{f}{x}\dd{x}{t} + \dd{f}{y}\dd{y}{t}\right)(t) \\ &= \inner{ \thevector{\dd{f}{x}(x(t), y(t)), \dd{f}{y}(x(t), y(t))} } { \thevector{\dd{x}{t}(t), \dd{y}{t}(t)} } \\ &= \inner{\nabla f(\gamma(t))}{\dd{\gamma}{t} (t)}. \end{align*}\]

  From above, we know that this is zero. Now note that we have \[ \dd{g}{t}(t_0) = \inner{\nabla f(\gamma(t_0))}{\dd{\gamma}{t}(t_0)} = \inner{\nabla f(\vector p)}{\gamma'(t_0)} \] but by the previous statement, \(\dd{g}{t}(t_0) = 0\), which exacty says that the gradient of \(f\) is orthogonal to \(\gamma\) at \(\vector p\). But \(\vector p\) was an arbitrary point on the level surface, and \(\gamma\) was an arbitrary curve through it. So \(\nabla f\) is orthogonal to every level curve through \(\vector p\), and this orthogonal to the tangent plane at \(\vector p\), and thus normal to the surface at \(\vector p\). Since \(\vector p\) was an arbitrary point on the level curve, this holds everywhere on the level curve, and for arbitrary level curves. Thus \(\nabla f\) is orthogonal to every level curve. \(\qed\)

Goal: Prove that for a continuously differentiable function \(f: \mathbb{R}^2 \to \mathbb{R}\), the gradient vector \(\nabla f(\mathbf{p})\) at any point \(\mathbf{p}\) on a level curve \(C = \{(x,y) \in \mathbb{R}^2 \mid f(x,y) = c\}\) is orthogonal to the tangent vector of every smooth curve in \(C\) passing through \(\mathbf{p}\).

<1>1. Let \(\mathbf{p} = (x_0, y_0) \in C\) such that \(f(\mathbf{p}) = c\). Let \(\gamma: (-\varepsilon, \varepsilon) \to \mathbb{R}^2\) be a differentiable parametrized curve such that \(\gamma(0) = \mathbf{p}\) and \(\gamma(t) \in C\) for all \(t \in (-\varepsilon, \varepsilon)\).

<1>2. The composite function \(g(t) = (f \circ \gamma)(t) = f(\gamma(t))\) is constant on \((-\varepsilon, \varepsilon)\) with value \(c\).

<1>3. \(g'(0) = 0\).

<1>4. By the multivariable Chain Rule, \(g'(0) = \nabla f(\gamma(0)) \cdot \gamma'(0) = \nabla f(\mathbf{p}) \cdot \gamma'(0)\).

<1>5. \(\nabla f(\mathbf{p}) \cdot \gamma'(0) = 0\).

<1>6. Conclusion: \(\nabla f(\mathbf{p})\) is orthogonal to the tangent vector \(\gamma'(0)\) of any curve lying on the level set through \(\mathbf{p}\).

A function \(f: \RR^2 \to \RR\) is totally differentiable at \(\vector x\) if there exists a linear map \(D: \RR^2 \to \RR\) such that \[f(\vector x + \vector h)- f(\vector x) = D(\vector x)\vector h + \varepsilon(\vector h) \text{ where } \frac{\varepsilon(\vector h)}{\norm{\vector h}} \to 0. \] Equivalently, it is when the following limit exists and satisfies \[ \lim_{\vector h \to \vector 0} \abs{\frac{f(\vector x + \vector h) - f(\vector x) - D(\vector x)\vector h}{\norm{\vector h}}} = 0, \] in which case we write \(f' = D\), or in this special case, \(f' = \nabla f\).

Substituting in $\vector x = \vector 0$, being differentiable at zero requires that $\exists D$ such that
$$
f(\vector h) - f(\vector 0) = D(\vector 0)\vector h + \varepsilon(\vector h), ~\frac{\varepsilon(\vector h)}{\norm{\vector h}} \to 0, 
\\ \text{ or } \\ 
\lim_{\vector h \to \vector 0} \abs{\frac{f(\vector h) - f(\vector 0) - D(\vector 0)\vector h}{\norm{\vector h}}} = 0.
$$
We'll make use the mean value theorem. The general statement is
$$
f' \text{ exists and continuous on } (a,b)
\implies \exists A \in [a,b] \suchthat \\
f(b) - f(a) = f'(c)(b-a)
$$
and here we'll use
$$
f_x \text{ exists and continuous on } (0, t) 
\implies \exists A \in [0, t] \suchthat \\ 
f(t, y) - f(0, y) = f_x(A, y)(t - 0) = tf_x(A, y) \\
\implies \color{purple} f(t, y) - f(0, y) = tf_x(A, y)
$$
where moreover 
$$
\lim_{t\to 0} f_x(A, y) = f_x(\lim_{t\to 0} A, y) = f(0, y)
$$ 
since $f_x$ was assumed to be continuous.

We'll also use the "best linear approximation" definition of differentiability. In general, it states
$$
f' \text{ exists on } (a,b) \\
\implies f(x) = f(a) + f'(a)(x-a) + o(x-a),~ \frac{o(x-a)}{x-a} \to 0 \\
\implies f(x) - f(a) = f'(a)(x-a) + o(x-a),~ \frac{o(x-a)}{x-a} \to 0 \\
\implies f(b) - f(a) = f'(a)(b-a) + o(b-a),~ \frac{o(b-a)}{b-a} \to 0
$$
and here we'll take $a=0, b=t$, so $x-a = x$ and $b-a = t$, and use the variant
$$
f_y \text{ exists on } (0, t) \\ 
\implies f(x, t) - f(x, 0) = f_y(x, 0)(t - 0) + o(t),~ \frac{o(t)}{t} \to 0 \\
\implies \color{green}f(x, t) - f(x, 0) = tf_y(x,0) + o(t), ~\frac{o(t)}{t} \to 0.
$$
We can then write
$$\begin{align*}
f(h_1, h_2) - f(0, 0)
&= {\color{green}f(h_1, h_2) - f(h_1, 0)} + {\color{purple}f(h_1, 0) - f(0, 0)} \\
&= {\color{green}h_2f_y(0, 0) + o(h_2)} + {\color{purple}h_1f_x(A, 0)}\\
&= (h_1 f_x(0, 0) - {\color{red}h_1 f_x(0, 0)}) + {\color{purple}h_1f_x(A, 0)} + {\color{green}h_2f_y(0, 0) + o(h_2)} \\
&= h_1 f_x(0, 0) + {\color{green}h_2 f_y(0, 0)} + h_1({\color{purple}f_x(A,0)} - {\color{red}f_x(0,0)}) + {\color{green}o(h_2)}
\end{align*}$$

Now, since $f_x$ is continuous, ${\color{purple}f_x(A,0)} - {\color{red}f_x(0,0)} \to 0$ and so $h_1(f_x(A,0) - f_x(0,0)) = o(h_1)$. We thus have
$$\begin{align*}
f(h_1, h_2) - f(0, 0)
&= h_1 f_x(0, 0) + h_2 f_y(0, 0) + o(h_1) + o(h_2) \\
&= h_1 f_x(0, 0) + h_2 f_y(0, 0) + o(\norm{\vector h}) \\ \\
&\implies 
f(h_1, h_2) - f(0, 0) - h_1 f_x(0, 0) - h_2 f_y(0, 0) = o(\norm{\vector h})
\end{align*}$$

and so if we define 
$$D(\vector h) = h_1 f_x(0, 0) + h_2 f_y(0, 0),
$$ 

we can take absolute values to find that 
$$
\abs{f(\vector h) - f(\vector 0) - D(\vector h)} = o(\norm{\vector h}) \\
\text{ and so } \\
\lim_{\vector h \to \vector 0} \abs{\frac{f(\vector h) - f(\vector 0) - D(\vector h)}{\norm{\vector h}}} = 0
$$

which forces $f$ to be totally differentiable at $\vector 0$ with derivative $D$.

Goal: Prove that if the partial derivatives \(f_x\) and \(f_y\) of \(f: \mathbb{R}^2 \to \mathbb{R}\) exist in an open neighborhood of \((0,0)\) and \(f_x\) is continuous at \((0,0)\), then \(f\) is (totally) differentiable at \((0,0)\) provided \(f_y(0,0)\) exists.

<1>1. Definition: \(f\) is differentiable at \(\mathbf{0} = (0,0)\) with total derivative \(L = (f_x(\mathbf{0}), f_y(\mathbf{0}))\) if and only if \[\lim_{\mathbf{h} \to \mathbf{0}} \frac{|f(h_1, h_2) - f(0,0) - h_1 f_x(0,0) - h_2 f_y(0,0)|}{\|\mathbf{h}\|} = 0.\]

<1>2. Decomposition of the difference: For small \(\mathbf{h} = (h_1, h_2)\), \[f(h_1, h_2) - f(0,0) = [f(h_1, h_2) - f(0, h_2)] + [f(0, h_2) - f(0,0)].\]

<1>3. Application of the single-variable Mean Value Theorem: There exists \(\theta \in (0, 1)\) such that \(f(h_1, h_2) - f(0, h_2) = h_1 f_x(\theta h_1, h_2)\).

<1>4. Linear approximation along the \(y\)-axis: \(f(0, h_2) - f(0,0) = h_2 f_y(0,0) + r(h_2)\) where \(\lim_{h_2 \to 0} \frac{r(h_2)}{h_2} = 0\).

<1>5. Bounding the remainder: \[|f(h_1, h_2) - f(0,0) - h_1 f_x(0,0) - h_2 f_y(0,0)| \le |h_1| |f_x(\theta h_1, h_2) - f_x(0,0)| + |r(h_2)|.\]

<1>6. The limit as \(\|\mathbf{h}\| \to 0\) vanishes: Proof: <2>1. Since \(|h_1| \le \|\mathbf{h}\|\), \(\frac{|h_1| |f_x(\theta h_1, h_2) - f_x(0,0)|}{\|\mathbf{h}\|} \le |f_x(\theta h_1, h_2) - f_x(0,0)|\). <2>2. As \(\mathbf{h} \to \mathbf{0}\), \((\theta h_1, h_2) \to (0,0)\). Since \(f_x\) is continuous at \((0,0)\), \(\lim_{\mathbf{h} \to \mathbf{0}} |f_x(\theta h_1, h_2) - f_x(0,0)| = 0\). <2>3. Since \(|h_2| \le \|\mathbf{h}\|\), \(\frac{|r(h_2)|}{\|\mathbf{h}\|} \le \frac{|r(h_2)|}{|h_2|} \to 0\) as \(h_2 \to 0\) (for \(h_2 \neq 0\); for \(h_2 = 0\), \(r(h_2) = 0\)). <2>4. Combining these bounds shows the quotient in <1>1 tends to \(0\).

<1>7. Conclusion: \(f\) is totally differentiable at \((0,0)\).

The region is \(0\le y\le x\le1\). Reversing the order of integration, \[ \int_0^1\int_y^1\sin(x^2)\,dx\,dy =\int_0^1\int_0^x\sin(x^2)\,dy\,dx =\int_0^1x\sin(x^2)\,dx. \] With \(u=x^2\), \[ \int_0^1x\sin(x^2)\,dx =\frac12\int_0^1\sin u\,du =\frac{1-\cos1}{2} =\boxed{\sin^2\!\left(\frac12\right)}. \]

\(\nabla f = (y, x)\). Along a level curve \(f(x,y) = c\) one may take a parametrization \(\gamma\) with \(f(\gamma(t)) = c\), so \(\frac{d}{dt}f(\gamma(t)) = 0\). By the chain rule this inner product is \(\langle \nabla f(\gamma(t)), \gamma'(t)\rangle\), hence \(\nabla f\) is orthogonal to every tangent of the level curve.

<1>1. Part (a): Definition of differentiability: <2>1. A function \(f: \mathbb{R}^n \to \mathbb{R}\) is differentiable at a point \(p \in \mathbb{R}^n\) if there exists a linear transformation \(L: \mathbb{R}^n \to \mathbb{R}\) (or equivalently a gradient vector \(\nabla f(p) \in \mathbb{R}^n\) such that \(L(h) = \langle \nabla f(p), h \rangle\)) satisfying: \[ \lim_{h \to 0} \frac{|f(p + h) - f(p) - L(h)|}{\|h\|} = 0. \] Equivalently, \(f(p + h) = f(p) + L(h) + o(\|h\|)\) as \(\|h\| \to 0\).

<1>2. Part (b): Non-differentiability of \(f(x, y)\) at \((0, 0)\): <2>1. If a function \(f\) is differentiable at \(p\), then \(f\) is continuous at \(p\).

<2>2. We evaluate the limit of \(f(x, y) = \frac{xy}{x^2 + y^2}\) as \((x, y) \to (0, 0)\) along lines \(y = mx\): For \(x \neq 0\): \[ f(x, mx) = \frac{x(mx)}{x^2 + (mx)^2} = \frac{m x^2}{x^2(1 + m^2)} = \frac{m}{1 + m^2}. \]

<2>3. Along the line \(y = 0\) (\(m = 0\)), \(\lim_{x \to 0} f(x, 0) = 0\). Along the line \(y = x\) (\(m = 1\)), \(\lim_{x \to 0} f(x, x) = \frac{1}{1 + 1^2} = \frac{1}{2} \neq f(0, 0)\). Since the limit depends on the path of approach, \(\lim_{(x, y) \to (0, 0)} f(x, y)\) does not exist. Thus \(f\) is discontinuous at \((0, 0)\).

<2>4. Since \(f\) is not continuous at \((0, 0)\), by <2>1 \(f\) cannot be differentiable at \((0, 0)\).

<1>3. Conclusion: \(f\) is defined to be differentiable if its linear approximation error is \(o(\|h\|)\), and \(f(x, y) = \frac{xy}{x^2+y^2}\) is not differentiable at \((0, 0)\) due to discontinuity. Q.E.D.

<1>1. Definition of the Jacobian matrix: <2>1. Let \(f: \mathbb{R}^n \to \mathbb{R}^m\) be given by coordinate functions \(f(x) = (f_1(x), f_2(x), \dots, f_m(x))^T\) for \(x = (x_1, \dots, x_n) \in \mathbb{R}^n\).

<2>2. If the partial derivatives \(\frac{\partial f_i}{\partial x_j}(a)\) exist at \(a \in \mathbb{R}^n\) for all \(1 \le i \le m\) and \(1 \le j \le n\), the Jacobian matrix \(J_f(a) \in M_{m \times n}(\mathbb{R})\) is defined as: \[ J_f(a) = \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(a) & \frac{\partial f_1}{\partial x_2}(a) & \cdots & \frac{\partial f_1}{\partial x_n}(a) \\ \frac{\partial f_2}{\partial x_1}(a) & \frac{\partial f_2}{\partial x_2}(a) & \cdots & \frac{\partial f_2}{\partial x_n}(a) \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1}(a) & \frac{\partial f_m}{\partial x_2}(a) & \cdots & \frac{\partial f_m}{\partial x_n}(a) \end{pmatrix}. \]

<1>2. Compute the Jacobian matrix for \(f(r, \theta) = (r\cos\theta, r\sin\theta)\): <2>1. The component functions are \(f_1(r, \theta) = r\cos\theta\) and \(f_2(r, \theta) = r\sin\theta\).

<2>2. Compute the partial derivatives with respect to \(r\) and \(\theta\): \[ \frac{\partial f_1}{\partial r} = \cos\theta, \qquad \frac{\partial f_1}{\partial \theta} = -r\sin\theta, \] \[ \frac{\partial f_2}{\partial r} = \sin\theta, \qquad \frac{\partial f_2}{\partial \theta} = r\cos\theta. \]

<2>3. The general Jacobian matrix at \((r, \theta)\) is: \[ J_f(r, \theta) = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}. \]

<1>3. Evaluation at \(a = (0, 0)\): <2>1. Setting \((r, \theta) = (0, 0)\): - \(\cos(0) = 1\), - \(-0 \cdot \sin(0) = 0\), - \(\sin(0) = 0\), - \(0 \cdot \cos(0) = 0\).

<2>2. Thus: \[ J_f(0, 0) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. \]

<1>4. Q.E.D.

Goal: State the multivariable chain rule and prove that level curve tangent lines are orthogonal to the gradient vector \(\nabla f(P)\).

<1>1. Part (a): Multivariable Chain Rule: Proof: <2>1. Let \(f: \mathbb{R}^2 \to \mathbb{R}\) and \(g: \mathbb{R} \to \mathbb{R}^2\) given by \(g(t) = (x(t), y(t))\) be differentiable functions. <2>2. The derivative of the composite function \(F(t) = (f \circ g)(t) = f(x(t), y(t))\) is given by: \[\frac{dF}{dt}(t) = \frac{\partial f}{\partial x}(x(t), y(t)) \frac{dx}{dt}(t) + \frac{\partial f}{\partial y}(x(t), y(t)) \frac{dy}{dt}(t).\] <2>3. In vector notation using the gradient vector \(\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\) and velocity vector \(g'(t) = (x'(t), y'(t))\): \[\frac{d}{dt}(f(g(t))) = \nabla f(g(t)) \cdot g'(t) = \langle \nabla f(g(t)), g'(t) \rangle.\]

<1>2. Part (b): Perpendicularity of \(\nabla f(P)\) to the Tangent Line: Proof: <2>1. Parameterize the graph of the implicit function \(y = h(x)\) as a smooth curve \(\gamma: I \to \mathbb{R}^2\): \[\gamma(x) = (x, h(x)).\] <2>2. The point \(P = (x_0, y_0) = \gamma(x_0)\). <2>3. The tangent vector to the graph at \(P\) is given by the derivative: \[\mathbf{T} = \gamma'(x_0) = \left( 1, h'(x_0) \right).\] <2>4. Since the graph lies entirely on the level set \(f(x, y) = f(P) = c\) for all \(x\) in a neighborhood of \(x_0\): \[f(\gamma(x)) = f(x, h(x)) = c \quad \text{for all } x \in I.\] <2>5. Differentiating both sides of the identity \(f(\gamma(x)) = c\) with respect to \(x\) using the Chain Rule from Part (a): \[\frac{d}{dx} [f(\gamma(x))] = \nabla f(\gamma(x)) \cdot \gamma'(x) = 0.\] <2>6. Evaluating at \(x = x_0\) where \(\gamma(x_0) = P\): \[\nabla f(P) \cdot \gamma'(x_0) = \nabla f(P) \cdot \mathbf{T} = 0.\] <2>7. Alternatively, by implicit differentiation: \[\frac{\partial f}{\partial x}(P) \cdot 1 + \frac{\partial f}{\partial y}(P) \cdot h'(x_0) = 0 \implies h'(x_0) = -\frac{f_x(P)}{f_y(P)} \quad (\text{if } f_y(P) \ne 0).\] Then: \[\nabla f(P) \cdot \mathbf{T} = (f_x(P), f_y(P)) \cdot \left( 1, -\frac{f_x(P)}{f_y(P)} \right) = f_x(P) - f_x(P) = 0.\] <2>8. Since the dot product \(\nabla f(P) \cdot \mathbf{T} = 0\), the gradient \(\nabla f(P)\) is perpendicular (orthogonal) to the tangent vector \(\mathbf{T}\) of the level curve at \(P\).

<1>3. Conclusion: The chain rule gives \(\frac{d}{dt} f(g(t)) = \nabla f(g(t)) \cdot g'(t)\); differentiating the level curve identity \(f(x, h(x)) = c\) yields \(\nabla f(P) \cdot (1, h'(x_0)) = 0\). Q.E.D.

Continuity of \(f_x\) and \(f_y\) at \((0,0)\) presupposes that both partial derivatives exist on some open disc \(B\) about \((0,0)\). Fix a direction \(\mathbf u=(u_1,u_2)\in\mathbb R^2\).

<1>1. For every sufficiently small \(h\neq0\) there are points \(\xi_h\) between \(0\) and \(hu_1\) and \(\eta_h\) between \(0\) and \(hu_2\) such that \[ f(hu_1,hu_2)-f(0,0)=f_x(\xi_h,hu_2)\,hu_1+f_y(0,\eta_h)\,hu_2. \]

<1>2. \(D_{\mathbf u}f(0,0)\) exists and equals \(f_x(0,0)u_1+f_y(0,0)u_2\).

<1>3. Q.E.D.

Let \[ g(x,y)=x^2+4y^2. \] At an extremum on \(g=8\), Lagrange multipliers give \[ (y,x)=\lambda(2x,8y), \] so \[ y=2\lambda x,\qquad x=8\lambda y. \] Neither \(x\) nor \(y\) can be zero at such a point, so substitution gives \[ 1=16\lambda^2. \] Hence \(\lambda=\pm1/4\).

If \(\lambda=1/4\), then \(y=x/2\). The constraint becomes \[ x^2+4(x/2)^2=2x^2=8, \] so \(x=\pm2\) and \(y=\pm1\) with matching signs. Thus \(xy=2\).

If \(\lambda=-1/4\), then \(y=-x/2\), giving the points \((2,-1)\) and \((-2,1)\) and hence \(xy=-2\).

Therefore the maximum value is \(2\) and the minimum value is \(-2\).

Goal: Evaluate the closed line integral \(\oint_\Delta y\,dx + 2x\,dy\) around the oriented triangle using Green’s Theorem.

<1>1. Setting up Green’s Theorem: Proof: <2>1. Let \(D \subset \mathbb{R}^2\) be the filled triangular region enclosed by the positively oriented (counterclockwise) boundary curve \(\partial D = \Delta\). <2>2. The vector field components are: \[P(x, y) = y, \qquad Q(x, y) = 2x.\] <2>3. Both \(P\) and \(Q\) have continuous partial derivatives on \(\mathbb{R}^2\): \[\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2x) = 2, \qquad \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y) = 1.\] <2>4. By Green’s Theorem: \[\oint_{\Delta} P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy = \iint_D (2 - 1) \, dA = \iint_D 1 \, dA = \operatorname{Area}(D).\]

<1>2. Computation of the Area of the Triangular Region \(D\): Proof: <2>1. The vertices of the triangle are \(A = (0, 1)\), \(B = (2, 0)\), and \(C = (2, 1)\). <2>2. Notice that the side \(AC\) is the horizontal line segment from \((0, 1)\) to \((2, 1)\) of length \(2\). <2>3. The side \(BC\) is the vertical line segment from \((2, 0)\) to \((2, 1)\) of height \(1\). <2>4. Since \(AC\) is horizontal and \(BC\) is vertical, the angle at vertex \(C = (2, 1)\) is a right angle (\(90^\circ\)). <2>5. Thus \(D\) is a right-angled triangle with base \(b = 2\) and height \(h = 1\). <2>6. The area of \(D\) is: \[\operatorname{Area}(D) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 1 = 1.\]

<1>3. Alternative Direct Line Integral Computation: Proof: <2>1. Path 1 (\(A(0,1) \to B(2,0)\)): Parameterize \(r_1(t) = (2t, 1 - t)\) for \(t \in [0, 1]\). - \(dx = 2\,dt\), \(dy = -dt\). - \(\int_{C_1} y\,dx + 2x\,dy = \int_0^1 (1 - t)(2) + 2(2t)(-1) \, dt = \int_0^1 (2 - 2t - 4t) \, dt = \int_0^1 (2 - 6t) \, dt = 2 - 3 = -1\). <2>2. Path 2 (\(B(2,0) \to C(2,1)\)): Parameterize \(r_2(t) = (2, t)\) for \(t \in [0, 1]\). - \(dx = 0\), \(dy = dt\). - \(\int_{C_2} y\,dx + 2x\,dy = \int_0^1 2(2) \, dt = 4\). <2>3. Path 3 (\(C(2,1) \to A(0,1)\)): Parameterize \(r_3(t) = (2 - 2t, 1)\) for \(t \in [0, 1]\). - \(dx = -2\,dt\), \(dy = 0\). - \(\int_{C_3} y\,dx + 2x\,dy = \int_0^1 (1)(-2) \, dt = -2\). <2>4. Summing all three edges: \[I = (-1) + 4 + (-2) = 1.\]

<1>4. Conclusion: The value of the line integral is \(1\). Q.E.D.

The differential form is exact: \[ x\,dy+y\,dx=d(xy). \] Therefore its integral around any closed piecewise smooth curve is zero. Since \(\partial R\) is closed, \[ \int_{\partial R}x\,dy+y\,dx=0. \] Equivalently, Green’s theorem gives the same result because \[ \frac{\partial x}{\partial x}-\frac{\partial y}{\partial y}=1-1=0. \]

<1>1. Exact differential formulation: <2>1. Consider the differential 1-form \(\omega = y\,dx + x\,dy\). By the product rule for differentials: \[ \omega = d(xy). \] <2>2. Thus \(\omega\) is an exact 1-form on \(\mathbb{R}^2\) with potential function \(f(x, y) = xy\).

<1>2. Application of the Fundamental Theorem of Line Integrals: <2>1. Let \(\tau\) be any piecewise smooth path starting at \(A = (1, 0)\) and ending at \(B = (3/5, 4/5)\). By the Fundamental Theorem for Line Integrals: \[ \int_\tau (x\,dy + y\,dx) = \int_\tau d(xy) = f(B) - f(A). \] <2>2. Evaluate the potential function at the endpoints: - \(f(A) = f(1, 0) = 1 \cdot 0 = 0\), - \(f(B) = f\left(\frac{3}{5}, \frac{4}{5}\right) = \frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25}\). <2>3. Therefore: \[ \int_\tau (x\,dy + y\,dx) = \frac{12}{25} - 0 = \frac{12}{25}. \]

<1>3. Alternative verification via trigonometric parametrization: <2>1. Parametrize the arc \(\tau\) by \(x(t) = \cos t, y(t) = \sin t\) for \(t \in [0, t_0]\), where \(\cos t_0 = 3/5\) and \(\sin t_0 = 4/5\). Then \(dx = -\sin t\,dt\) and \(dy = \cos t\,dt\). <2>2. The integrand becomes: \[ x\,dy + y\,dx = (\cos^2 t - \sin^2 t)\,dt = \cos(2t)\,dt. \] <2>3. Integrating: \[ \int_0^{t_0} \cos(2t)\,dt = \left[ \frac{\sin(2t)}{2} \right]_0^{t_0} = \sin(t_0)\cos(t_0) = \frac{4}{5} \cdot \frac{3}{5} = \frac{12}{25}. \]

<1>4. Conclusion: \(\int_\tau x\,dy + y\,dx = \frac{12}{25}\). Q.E.D.

<1>1. The value depends on the orientation of \(C\). For the positive (counterclockwise) orientation, \[ \int_C 2x\,dx+x^2y\,dy=\frac12. \] For the clockwise orientation the value is \(-\frac12\). The source does not specify an orientation; the steps below compute the positively oriented integral, which is the orientation Green’s theorem uses.

<1>2. Compute the positively oriented integral directly by traversing the four sides \[ (0,0)\to(1,0)\to(1,1)\to(0,1)\to(0,0). \]

<1>3. The bottom side contributes \(1\).

<1>4. The right side contributes \(\frac12\).

<1>5. The top side contributes \(-1\), and the left side contributes \(0\).

<1>6. Therefore the direct computation gives \[ 1+\frac12-1+0=\frac12. \]

<1>7. Green’s theorem gives the same value.

<1>8. Reversing the orientation changes the sign of every line integral, giving \(-\frac12\) for clockwise orientation.

Part (a).

<1>1. If \(\gamma: [a,b] \to \RR^2\) is a piecewise-smooth path with \(\gamma(t) = (x(t), y(t))\), then \[\int_\gamma f\,dx + g\,dy = \int_a^b \left[ f(x(t), y(t))\, x'(t) + g(x(t), y(t))\, y'(t) \right] dt.\]

Part (b).

<1>1. Parametrize \(\gamma(t) = (1-t, t)\) for \(t \in [0,1]\).

<1>2. \(x'(t) = -1\) and \(y'(t) = 1\).

<1>3. The integrand is \(f\,x' + g\,y' = 2xy(-1) + (x^2 + y^2)(1) = -2(1-t)t + (1-t)^2 + t^2\).

<1>4. \(\int_\gamma f\,dx + g\,dy = \int_0^1 \left[ -2t(1-t) + (1-t)^2 + t^2 \right] dt = \int_0^1 (1 - 4t + 4t^2)\, dt\).

<1>5. \(\int_0^1 (1 - 4t + 4t^2)\, dt = \left[ t - 2t^2 + \tfrac{4}{3}t^3 \right]_0^1 = 1 - 2 + \tfrac{4}{3} = \tfrac{1}{3}\).

<1>6. Hence \(\int_\gamma f\,dx + g\,dy = \tfrac{1}{3}\).

Part (c).

<1>1. No such path \(\beta\) exists. <2>1. The form \(f\,dx + g\,dy\) is exact.

<2>2. The line integral of an exact form depends only on the endpoints.

<2>3. Hence every path from \(P\) to \(Q\) gives the same value \(\tfrac{1}{3}\).

<1>2. Q.E.D.

<1>1. The path \(\gamma\) is the positively oriented boundary of the quarter disk \[ D=\{(x,y):x\ge0,\ y\ge0,\ x^2+y^2\le4\}. \]

<1>2. Write the integrand as \[ P\,dx+Q\,dy \quad\text{with}\quad P(x,y)=y,\qquad Q(x,y)=2x. \] Then \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2-1=1. \]

Parametrize the line segment by \[ r(t)=(1-t)(2,1)+t(3,5)=(2+t,1+4t),\qquad 0\leq t\leq1. \] Then \[ \int_C(2x+y)\,dx+y\,dy =\int_0^1(9+22t)\,dt =20. \]

For \(F=(P,Q)=(2x+y,y)\), path independence on \(\mathbb R^2\) would require \(P_y=Q_x\). Here \(P_y=1\) and \(Q_x=0\), so the integral is path dependent.

Goal: Prove that the vector field \(\mathbf{F}(x, y) = (x + y^3)\mathbf{i} + (e^y + 3xy^2)\mathbf{j}\) is conservative on \(\mathbb{R}^2\) by explicitly finding a scalar potential function \(f(x, y)\), thereby establishing path-independence.

<1>1. Construction of a potential function \(f(x, y)\): Proof: <2>1. Let \(P(x, y) = x + y^3\) and \(Q(x, y) = e^y + 3xy^2\) be the components of the 1-form \(\omega = P\,dx + Q\,dy\). <2>2. We seek a \(C^1\) function \(f(x, y)\) on \(\mathbb{R}^2\) such that \(\nabla f = (P, Q)\), which requires: \[\frac{\partial f}{\partial x} = x + y^3, \qquad \frac{\partial f}{\partial y} = e^y + 3xy^2.\] <2>3. Integrating \(\frac{\partial f}{\partial x} = x + y^3\) with respect to \(x\) while holding \(y\) constant yields \[f(x, y) = \int (x + y^3)\,dx = \frac{1}{2}x^2 + xy^3 + g(y),\] where \(g(y)\) is an arbitrary differentiable function of \(y\). <2>4. Differentiating this expression with respect to \(y\): \[\frac{\partial f}{\partial y} = 3xy^2 + g'(y).\] <2>5. Setting \(\frac{\partial f}{\partial y} = Q(x, y) = e^y + 3xy^2\) gives \(3xy^2 + g'(y) = e^y + 3xy^2\), so \(g'(y) = e^y\). <2>6. Integrating with respect to \(y\) gives \(g(y) = e^y + C\). Setting \(C = 0\) yields the potential function \[f(x, y) = \frac{1}{2}x^2 + xy^3 + e^y.\] <2>7. Direct verification: \(\frac{\partial f}{\partial x} = x + y^3 = P(x, y)\) and \(\frac{\partial f}{\partial y} = 3xy^2 + e^y = Q(x, y)\), so \(\nabla f = (P, Q)\) on all of \(\mathbb{R}^2\).

<1>2. Path-independence via the Fundamental Theorem for Line Integrals: Proof: <2>1. Let \(C\) be any piecewise smooth curve in \(\mathbb{R}^2\) parameterized by \(\mathbf{r}(t) = (x(t), y(t))\) for \(a \le t \le b\), with initial point \(A = \mathbf{r}(a)\) and terminal point \(B = \mathbf{r}(b)\). <2>2. By the chain rule: \[\int_C (x + y^3)\,dx + (e^y + 3xy^2)\,dy = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt = \int_a^b \frac{d}{dt}[f(\mathbf{r}(t))]\,dt.\] <2>3. By the Fundamental Theorem of Calculus: \[\int_C (x + y^3)\,dx + (e^y + 3xy^2)\,dy = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B) - f(A).\] <2>4. The value of the line integral depends solely on the endpoints \(A\) and \(B\), and is completely independent of the path \(C\) connecting them.

<1>3. Conclusion: Proof: The line integral is path-independent.

We first note that we can rewrite the equation of the region to obtain something more familiar: \(x^2 + y^2 = 2x \implies (x-1)^2 + y^2 = 1\), which is a translated circle. Integrating over this region will be easy compared to the line integral, so we apply Green’s theorem: \[ \int_C xe^x ~dx + ye^y +x^2 ~dy = \iint_D 2x ~dA. \]

We can parameterize this region as \[ D = \theset{x^2+y^2-2x = 0 \suchthat (x,y) \in \RR^2, y \geq 0} = \theset{(r(1+ \cos\theta), r\sin\theta) \suchthat \theta \in [0, \pi), r\in [0, 1]}. \]

Noting that \(dA = r~dr~d\theta\), we can then integrate \[ \iint_D 2x ~dA = \int_0^{\pi} \int_0^1 2(r(1 + \cos\theta)) r ~dr ~d\theta \\ = 2\int_0^{\pi} \int_0^1 r^2(1+\cos \theta) ~dr ~d\theta \\ = 2\int_0^{\pi} \frac 1 3 r^3(1+\cos \theta) \bigg\rvert_0^1 ~d\theta\\ = \frac 2 3 \int_0^{\pi} (1+\cos \theta) ~d\theta\\ = \frac 2 3 ( \theta + \sin \theta) \bigg\rvert_0^\pi \\ = \frac 2 3 [(\pi + 0) - (0 + 0)] = \frac 2 3 \pi. \qed \]

Let \[ P(x,y)=\frac{-y}{x^2+y^2}, \qquad Q(x,y)=\frac{x}{x^2+y^2}. \] Away from the origin, \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0. \]

If the origin lies outside \(C\), then \(P,Q\) are continuously differentiable on the disk bounded by \(C\). Green’s theorem therefore gives \[ \oint_C P\,dx+Q\,dy=0. \]

If the origin lies inside \(C\), choose a small positively oriented circle \(C_\varepsilon\) centered at the origin and contained inside \(C\). Apply Green’s theorem to the annular region between \(C\) and \(C_\varepsilon\). Its positively oriented boundary is \(C-C_\varepsilon\), so \[ \oint_C P\,dx+Q\,dy-\oint_{C_\varepsilon}P\,dx+Q\,dy=0. \] Parametrize \(C_\varepsilon\) by \[ x=\varepsilon\cos t,\qquad y=\varepsilon\sin t,\qquad 0\le t\le2\pi. \] Then \[ P\,dx+Q\,dy =\sin^2t\,dt+\cos^2t\,dt=dt. \] Hence \[ \oint_C P\,dx+Q\,dy =\int_0^{2\pi}dt =2\pi. \]