Let : DCR² → R³ be a nice parametrization of surface S. We introduce the following notation:(a) Show thatE = ||T,||²,F=T₁ Tv, G = ||T,||².||Tux Tv||= √√EG - F².This gives an alternative formula for the surface area:A(S) =VEG-F2dSWhat happens to this formula if T and T₁, are orthogonal?(b) The sphere of radius R is parametrized by(u,v) = (Rsin(u) cos(v), R sin(u) sin(v), Rcos(u)), 0 ≤ u≤π, 0 ≤v≤2π.Use the formula above to find its surface area.(c) Determine the surface area of the portion of x+4y+8z = 4 that is inside the cylinder x² + y² = 16.

Certainly! Here are the final answers for each part:(a) The formula ∣∣Tu×Tv∣∣2=EG−F2provides an…

Answered: Let : DCR² → R³ be a nice…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Let S be the surface in R³ defined by the equation x4. y³ – 5z = 8. Find a real-valued function g (x, y) of two variables such that S is the graph of g. (Express numbers in exact form. Use symbolic notation and fractions where needed.) g(x, y) = =
Which of the following statements is NOT correct? (A) The function u(x, y) = x² - y² is harmonic. (B) The function f(2)=3 is analytic on the disk of center 0 and radius 1. (C) The function cosh(z) is entire. (D) The function |z|2 is entire. (E) fedz = 0 where C is the counterclockwise unit circle.
Q #1 Let C denote the positively oriented boundary of the square whose sides lie along the lines x = ±2 and cosh z Edz . y =+ 2. Compute |
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Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0) (a) Find a piecewise smooth parametrization of the path C. (t,0) (2-t,t - 1) (0,3 – t) r(t) = (b) Evaluate √(x + 4√Y) ds. 0 ≤t≤ 1 1 ≤t≤2 2 ≤ t ≤ 3
Consider F and C below. F(x, y) = x2 i + y? i C is the arc of the parabola y = 3x2 from (-2, 12) to (-1, 3) Exercise (a) Find a function f such that F = Vf. Step 1 If Vf(x, y) = F(x, y) = x²i + y²j, then 2. fx(x, y) = and fy(x, y) = y2 Step 2 Since f,(x, y) = x², find f(x, y). 3 f(x, y) = 3 + g(y) 3 Step 3 Now, we know the following. f,(x, y) = ду + g»)] = 9 (|
Consider surface z = sin(T(3x+ 2y)) + x²y².
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Consider F and C below. F(x, y) = x2 i + y² j C is the arc of the parabola y = 5x2 from (1, 5) to (2, 20) Exercise (a) Find a function f such that F = Vf. Step 1 If Vf(x, y) = F(x, y) = x²i + y²j, then f(x, y) = and f,(x, y) = y2 y? Step 2 Since f(x, y) = x², find f(x, y). f(x, y) = 3 + g(y) Step 3 Now, we know the following. fy(x, y) = Step 4 Since we must have f,(x, y) = y² = g'(y), then v° g(y) = 3 y3 Substituting this back into f(x, y), we have f(x, y) = 3 3 3 Exercise (b) Use part (a) to evaluate Vf · dr along the given curve C. Step 1 Since we want a specific potential function, we will choose K = 0, and so f(x, y) = | 1/3 . 1/3 V We know that if the curve C is given by r(t) with a sts b, then F. dr = f(r(b)) – f(r(a)). Step 2 Our curve C is the arc of the parabola y = 5x2 that begins at (1, 5) and ends at (2, 20). Therefore, we can conclude the following. F. dr = f(2, 20) – f(1, 5) - ( -2456 Submit Skip (you cannot come back)
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