Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface
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Chapter 15 Solutions
Calculus
- Fhd the equation of the tangent plane and normal line to the surface z = 2+y at the point (1, 1,3).arrow_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). -4i + 4k (b) Find an implicit equation for the tangent plane to the surface at (-2, 2, 4). X-Z +6=0arrow_forwardTangent of x?/3 + y2/3 + z2/3 = a²/3 surface at any point ( xo , Yo ,Zo ) Show that the sum of the squares of the intersecting axes of the plane is constant.arrow_forward
- Check 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_forwardSinx dA where R is the trangle in xy-plane bounded by the x-anise, the line y=x and. the line =arrow_forwardParameterize the intersection of the cone z = x2 + y2 and the plane z = 2x + 4y + 20. Find the tangent line at the point (4, -2, 20).arrow_forward
- 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_forwardCheck that the point (-1, 1, 2) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y, z), find à vector normal to the surface and an equation for the tangent plane to the surface at (-1, 1, 2). vector normal = tangent plane: z= 2x²-4y²+3z2 = 10arrow_forwardCheck that the point (1,-1,2) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y, z), find a vector normal to the surface and an equation for the tangent plane to the surface at (1,-1,2). vector normal = tangent plane: 4x² - y² +3z² = 15arrow_forward
- A normal line to a surface S at a point (x, y, z) E S is a line perpendicular to the tangent plane to S at (x, y, 2). a) Find the second intersection point between the normal line to the level surface F(x, y, z) y?- z2 = 1 at the point (1, –1,-1) and the same surface (normal line intersects the level surface at 2 points). x2 +arrow_forwardParameterize the portion of the surface f (x, y) =x² + y² that lies below the plane y +z = 1.arrow_forwardQ1,- A- Locate the centroid (X only) of the shaded area y=12 -3x 3 m Fig. L.A 2 marrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage