![]() F5=Īs in the case of Green's Theorem, if F is a vector field such that on a surface S with boundary curve C, then Stokes' Theorem says thatĬomputes the surface area of S. That lies below the plane, and F5 is the following input cell. Verify Stokes' theorem for the case in which S is the portion of the upper sheet of the hyperbolic paraboloid NdS=simplify(cross(diff(sigma,r),diff(sigma,t))) int(realdot(subs(F4,boundary),diff(boundary,t)),t,0,2*pi) Let us now evaluate both sides of Stokes' theorem in this case. This has the great advantage that we can parametrize the boundary curve by setting r to 2. We can parametrize S conveniently using polar coordinates. We will choose S to be the portion of the hyperbolic paraboloid that is contained in the cylinder, oriented by the upward normal n, and we will take F4 as defined below. Let us perform a calculation that illustrates Stokes' Theorem. Stokes' Theorem states that if S is an oriented surface with boundary curve C, and F is a vector field differentiable throughout S, then Where is the surface of Problem 1 and G is defined by the input line below. Fpar = subs(F,ellipsoid)įlux = symint2(realdot(Fpar,ndS),p,0,pi,t,0,2*pi) subs(ndS,)īefore we can evaluate the integral, we must evaluate F in terms of the parameters. The outward normal should point in the positive x direction. We can check that we have the outward normal by setting and, giving us the point of the ellipsoid that is on the positive x-axis. NdS = cross(diff(ellipsoid,p),diff(ellipsoid,t)) ![]() Where the integral is taken over the ellipsoid E of Example 1, F is the vector field defined by the following input line, and n is the outward normal to the ellipsoid. Flux IntegralsĪlso allows us to compute flux integrals over parametrized surfaces. Where is the surface whose area you found in part (a). (a) Compute the surface area of the portion of the paraboloid that lies above the -plane. This is right since we can also "unwrap" the cone to a sector of a circular disk, with radius and outer circumference (compared to for the whole circle), so the surface area is The picture is: ezsurf(cone(1),cone(2),cone(3),)Ĭonesurffactor=simple(veclength(cross(diff(cone,r),diff(cone,t)) )) ![]() In cylindrical coordinates the equation of the portion of the cone with is just, so we can take as our parameters r and t (representing theta). Here's another example: suppose we want the surface area of the portion of the cone between and. Integral=newnumint2(surffactor*func,p,0,pi,t,0,2*pi) Over the surface, we must express it in terms of the parameters and insert the result as a factor in the integrand. This is going to be too complicated to integrate symbolically, so we do a numerical integration, using newnumint2 from the numerical integration toolbox nit, which is a folder in the mfiles you downloaded on Assignment 1. Surffactor = simple(veclength(cross(diff(ellipsoid,t). Now we compute the surface area factor: realdot = u*transpose(v) To check that this really is a parametrization, we verify the original equation: simplify(subs((x^2/4)+(y^2/9)+z^2,ellipsoid))Īnd we can also draw a picture with ezsurf: ezsurf(ellipsoid(1),ellipsoid(2),ellipsoid(3),) We may parametrize this ellipsoid as we have done in the past, using modified spherical coordinates: syms x y z p tĮllipsoid= To see how this works, let us compute the surface area of the ellipsoid whose equation is Which enables us to compute the area of a parametrized surface, or to integrate any function along the surface with respect to surface area. Thus, taking lengths on both sides of the above formula above gives The key idea behind all the computations is summarized in the formulaĪre vectors, and their cross-product is a vector with two important properties: it is normal to the surface parametrized by r, and its length gives the scale factor between area in the parameter space and the corresponding area on the surface. In other words, the surface is given by a vector-valued function r (encoding the x, y, and z coordinates of points on the surface) depending on two parameters, say u and v. Recall that a surface is an object in 3-dimensional space that locally looks like a plane. We begin this lesson by studying integrals over parametrized surfaces.
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