Lateral Surface Area In Exercises 65–-72, find the area ofthe lateral surface (see figure) over the curve C in the x y-plane and under the surface
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Calculus: Early Transcendental Functions (MindTap Course List)
- Consider the surface What is the equation of the tangent plane to the surface at the point P(3,3,-2)? O 32³ +3zy+zyz² = 0 O 34z + 7y - 12z = 147 O 3x+3y-2z = 0 O 34z + 7y-12z = 0 O 3x+3y-2z=147 What is the equation of the line normal to the surface at the point P(3,3,-2) ? 3 -0-0)-C) 3 +t 21 -2 O O O NC O =t No wo wo -2 102 -0-0-0) 21 +t 3 -36 102 102 =1 21 -36 -36 3 2 3 -O-C-(C) <=8 3 +t 102 32³ +3zy+zyz² = 144. 21 -36arrow_forwardPlz complete solution with 100%accuracy I vill upvote if corrected otherwise downvotearrow_forwardCheck that the point (-2, 2, 4) lies on the surface cos(x + y) = exz+8 (a) View this surface as a level surface for a function f(x, y, z). Find a vector normal to the surface at the point (-2, 2, 4). (b) Find an implicit equation for the tangent plane to the surface at (-2, 2, 4).arrow_forward
- Sketch the surface x = 2y2 +3z2arrow_forwardTutorial Exercise Find the area of the part of the plane 3x + 2y + z = 6 that lies in the first octant. Part 1 of 5 2 The surface S is the graph of a function z = f(x, y) over a region D in the xy-plane. The surface area of S can be calculated by A(S) = az 1 + dz dA. ду f(x, y) = 6 – 3x az 2y. We, therefore, have ax əz -3 and -2 The plane 3x + 2y + z = 6 is the graph of the function z = 3.x ду Part 2 of 5 Therefore, =7 V1+(-3)² + (-2)² dA = V14 dA. Since we want the area of the section of the plane that lies above the first octant, then the region D will be the triangular region bounded by the x-axis, the y-axis, and the line formed by the intersection of the plane with the plane z = 0. Substituting z = 0 into 3x + 2y + z = 6 and solving for y, we find that this line of intersection has the equation y = x + Submit Skip (you cannot come back)arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage