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Calc-3 Discussion 08

Discussion Topics covered: Constrained optimization via Lagrange multipliers, geometric tangency of level curves and constraint boundaries, resolving non-linear algebraic systems without division by zero, shadow price and physical sensitivity analysis, iterated double integrals over rectangular domains, Fubini's Theorem, product separability, 2D spatial average value, double integrals over general planar regions, Type I vs. Type II geometries, curve intersection bounding, 4-step order-switching protocol, polar coordinate conversions and curve geometry, trigonometric power-reduction formulas, 3D traces and quadric surface projections

Part A: Review This Week's Lecture Topics (~30 mins)

Revisiting Constrained Optimization & The Method of Lagrange Multipliers

  • Formulating the fundamental geometric condition for constrained optimization: for an objective function f(x,y)f(x, y) subject to a constraint g(x,y)=kg(x, y) = k, extrema along the constraint occur where level curves of ff are tangent to the constraint curve g(x,y)=kg(x, y) = k
  • Enforcing gradient collinearity via the vector Lagrange multiplier equation:
    ∇f(x,y)=λ∇g(x,y)subject tog(x,y)=k\nabla f(x, y) = \lambda \nabla g(x, y) \quad \text{subject to} \quad g(x, y) = k
    where λ∈R\lambda \in \mathbb{R} is the scalar multiplier
  • Extending the method to 3D surfaces subject to two simultaneous constraints g(x,y,z)=kg(x, y, z) = k and h(x,y,z)=ch(x, y, z) = c (optimization along the space curve formed by the intersection of two surfaces):
    ∇f=λ∇g+μ∇hsubject tog(x,y,z)=k,h(x,y,z)=c\nabla f = \lambda \nabla g + \mu \nabla h \quad \text{subject to} \quad g(x, y, z) = k, \quad h(x, y, z) = c
  • Executing robust algebraic strategies to solve non-linear Lagrange systems without division by zero:
    • Decompose the vector equation into scalar components: fx=λgx, fy=λgy, fz=λgzf_x = \lambda g_x, \, f_y = \lambda g_y, \, f_z = \lambda g_z
    • Equate proportional ratios fxgx=fygy\frac{f_x}{g_x} = \frac{f_y}{g_y} only after explicitly checking and isolating cases where denominators vanish (gx=0g_x = 0 or gy=0g_y = 0)
    • Never divide both sides of an equation by a variable factor (e.g., x(2y−λ)=0  ⟹  x=0x(2y - \lambda) = 0 \implies x = 0 or λ=2y\lambda = 2y); cancelling xx discards valid critical points along the coordinate axes
  • Interpreting the Lagrange multiplier λ\lambda as a quantitative measure of physical system sensitivity and marginal rate of return (shadow pricing):
    λ=dfoptdk\lambda = \frac{df_{\text{opt}}}{dk}
    where λ\lambda represents the instantaneous change in the optimized objective value per unit relaxation of the constraint parameter kk

Double Integrals over Rectangular Regions & Spatial Averages

  • Interpreting double integrals as spatial volume accumulation: for a continuous function f(x,y)≥0f(x, y) \ge 0 over a rectangular domain R=[a,b]×[c,d]R = [a, b] \times [c, d], the double integral computes the exact volume under the surface z=f(x,y)z = f(x, y) above RR:
    ∬Rf(x,y) dA=lim⁡Δx→0Δy→0∑i=1m∑j=1nf(xij∗,yij∗) Δx Δy\iint_R f(x, y) \, dA = \lim_{\substack{\Delta x \to 0 \\ \Delta y \to 0}} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \, \Delta x \, \Delta y
    When f(x,y)=1f(x, y) = 1, the integral yields the planar area: Area(R)=∬R1 dA=(b−a)(d−c)\text{Area}(R) = \iint_R 1 \, dA = (b - a)(d - c)
  • Applying Fubini's Theorem on rectangular domains: because boundary limits are constants, integration order is freely interchangeable:
    ∬Rf(x,y) dA=∫cd(∫abf(x,y) dx)dy=∫ab(∫cdf(x,y) dy)dx\iint_R f(x, y) \, dA = \int_{c}^{d} \left(\int_{a}^{b} f(x, y) \, dx\right) dy = \int_{a}^{b} \left(\int_{c}^{d} f(x, y) \, dy\right) dx
  • Maintaining inner variable substitution integrity: when evaluating the inner integral with respect to yy, treat xx strictly as an algebraic constant; when integrating with respect to xx, treat yy as a constant; outer limits of integration must strictly be numeric constants
  • Exploiting the product separability property: if the integrand factors into independent single-variable functions f(x,y)=g(x)⋅h(y)f(x, y) = g(x) \cdot h(y), decouple the double integral into the product of two independent single integrals:
    ∬Rg(x)h(y) dA=(∫abg(x) dx)(∫cdh(y) dy)\iint_R g(x)h(y) \, dA = \left(\int_a^b g(x) \, dx\right) \left(\int_c^d h(y) \, dy\right)
  • Calculating the continuous spatial average value of a two-variable function across a planar region:
    favg=1Area(R)∬Rf(x,y) dAf_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R f(x, y) \, dA

Double Integrals over General Planar Regions (Type I & Type II Geometries)

  • Classifying and evaluating Type I Regions (Vertical Slicing): the domain is bounded between vertical constant lines x=ax = a and x=bx = b, with yy varying between continuous lower and upper boundary functions g1(x)≤y≤g2(x)g_1(x) \le y \le g_2(x):
    ∬Df(x,y) dA=∫ab∫g1(x)g2(x)f(x,y) dy dx\iint_D f(x, y) \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx
    Outer limits are constants [a,b][a, b]; inner limits are functions of the outer variable: ybottom=g1(x)y_{\text{bottom}} = g_1(x) to ytop=g2(x)y_{\text{top}} = g_2(x)
  • Classifying and evaluating Type II Regions (Horizontal Slicing): the domain is bounded between horizontal constant lines y=cy = c and y=dy = d, with xx varying between continuous left and right boundary functions h1(y)≤x≤h2(y)h_1(y) \le x \le h_2(y):
    ∬Df(x,y) dA=∫cd∫h1(y)h2(y)f(x,y) dx dy\iint_D f(x, y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy
    Outer limits are constants [c,d][c, d]; inner limits are functions of the outer variable: xleft=h1(y)x_{\text{left}} = h_1(y) to xright=h2(y)x_{\text{right}} = h_2(y)
  • Establishing outer constant integration limits by solving simultaneous boundary curve equations (g1(x)=g2(x)g_1(x) = g_2(x) or h1(y)=h2(y)h_1(y) = h_2(y)) to find boundary intersection coordinates
  • Partitioning compound multi-curve boundaries into disjoint sub-regions D=D1∪D2D = D_1 \cup D_2 whenever the upper or lower boundary changes functional definitions across the domain

Switching the Order of Integration

  • Recognizing when an iterated integral demands order reversal: when an inner single-variable integrand lacks an elementary antiderivative (e.g., ∫ex2 dx\int e^{x^2} \, dx, ∫sin⁡yy dy\int \frac{\sin y}{y} \, dy, ∫cos⁡(x2) dx\int \cos(x^2) \, dx, ∫1+y3 dy\int \sqrt{1 + y^3} \, dy), switching the order introduces an exterior variable multiplier (xx or yy) from the newly integrated boundary limits, unlocking direct uu-substitution
  • Executing the systematic 4-step order-switching protocol:
    1. Extract Limits: Translate the given iterated integral bounds into explicit inequalities defining the planar domain DD
    2. Sketch the Domain: Plot the boundary curves on the xyxy-plane, identifying all corner vertices and shading the enclosed region
    3. Invert Boundary Functions: Solve the inner boundary curves for the opposite variable (e.g., inverting y=g(x)y = g(x) into x=g−1(y)x = g^{-1}(y))
    4. Re-slice the Domain: Convert vertical slices to horizontal sweeps (or vice versa), establish new constant outer limits and variable inner limits, and evaluate the resulting integral

Part B: Next Week's Prerequisite Refresher (~30 mins)

Review the following prerequisite skills required for Double Integrals in Polar Coordinates (Session 15) and Triple Integrals (Session 16):

  • Polar Coordinate Conversions & Radial Geometry
    • Mapping points and position vectors between Cartesian coordinates (x,y)(x, y) and polar coordinates (r,θ)(r, \theta) using x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, r=x2+y2r = \sqrt{x^2+y^2}, and θ=arctan⁡(y/x)\theta = \arctan(y/x) with quadrant awareness
    • Transforming Cartesian boundary equations into polar form r=g(θ)r = g(\theta), converting circles centered at the origin (x2+y2=a2  ⟹  r=ax^2 + y^2 = a^2 \implies r = a), radial rays (y=mx  ⟹  θ=αy = mx \implies \theta = \alpha), and offset circles (x2+y2=2ax  ⟹  r=2acos⁡θx^2 + y^2 = 2ax \implies r = 2a\cos\theta)
    • Establishing angular sweep intervals [−π/2,π/2][-\pi/2, \pi/2] and [0,π][0, \pi] for circular domains tangent to the coordinate axes
  • Trigonometric Power-Reduction Formulas (Half-Angle Integration)
    • Applying cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1 + \cos(2\theta)}{2} and sin⁡2θ=1−cos⁡(2θ)2\sin^2\theta = \frac{1 - \cos(2\theta)}{2} to reduce quadratic trigonometric factors into directly integrable first-degree terms
    • Evaluating angular definite integrals over symmetric full rotations [0,2π][0, 2\pi] and quadrants [0,π/2][0, \pi/2], leveraging periodic cancellation ∫02πcos⁡(2θ) dθ=0  ⟹  ∫02πcos⁡2θ dθ=π\int_0^{2\pi} \cos(2\theta) \, d\theta = 0 \implies \int_0^{2\pi} \cos^2\theta \, d\theta = \pi
    • Resolving polar double integrals for moments of inertia and surface areas involving x2=r2cos⁡2θx^2 = r^2\cos^2\theta and y2=r2sin⁡2θy^2 = r^2\sin^2\theta
  • Traces and Cross-Sections of Surfaces (3D Shadow Projections)
    • Slicing 3D surfaces z=f(x,y)z = f(x, y) with coordinate planes (x=0,y=0,z=0x = 0, y = 0, z = 0) and parallel cutting planes to identify characteristic 2D intersection curves
    • Visualizing and classifying standard quadric surfaces (circular and elliptic paraboloids z=x2/a2+y2/b2z = x^2/a^2 + y^2/b^2, cones z=x2+y2z = \sqrt{x^2+y^2}, spheres x2+y2+z2=R2x^2+y^2+z^2 = R^2, and cylinders x2+y2=a2x^2+y^2 = a^2)
    • Projecting 3D solid boundaries onto the coordinate shadow plane DxyD_{xy} to establish floor surfaces z=u1(x,y)z = u_1(x, y), ceiling surfaces z=u2(x,y)z = u_2(x, y), and outer 2D integration limits for triple integrals ∭Ef(x,y,z) dV\iiint_E f(x, y, z) \, dV

Part C: General Q/A

Open the floor to questions from students regarding any topic