Use a parametrization to express the area of the surface as...

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. The portion of the cone z = 2??x^2 + y^2 between the planes z = 4 and z = 6 Let u = r and v = ? and use cylindrical coordinates to parametrize the surface. Set up the double integral to find the surface area. ? ? ( ) du dv (Type exact answers.) After evaluating the double integral, the surface area is . (Type an exact answer, using ? and radicals as needed.)

Transcript

00:01 So in this case, we are giving the cone.

00:04 So let's just sketch a cone here on the right side.

00:08 And between two planes, so here we have a plane at say z equals to four, and higher yet, we have a different plane up there.

00:19 Say at z equals four, six.

00:21 Then where is s? so s is described as the part of the cone that lies between the planes.

00:29 So we try to demonstrate where that is, which is this part of here.

00:34 This is s, the surface, which is the portion of the cone.

00:39 So s is part of the cone, z equals 2, 2 times the square of x squared plus y squared, where between z equals 4 and z equals 6.

00:58 So next, next, we want to parameterize this in terms of s in terms of r theta.

01:07 So that is represent this as r of r from a theta, where x would be r, cosine theta, while the r sine theta, and z will just be two times a square root of r squared, which is two times r.

01:22 So r from here is just going to be cosine theta, comma, sine theta, comma, two, and r of data in our case is just going to equal to r well they ask us to use of u and v instead of r but pretty much they are pretty much the same exact thing so we're going to have so let's say u equals r and data equals v we can do that in a minute so r of theta in our case is going to be negative r sine theta comma r cosine data comma zero and take the cross product of them so i put ij k on top of this then rso r times r of theta in this case the cross product so the i .t component is this and that which is zero and then is minus two r plus in theta and then the j hat component is negative two um let's let's see the j hat component is this time dot and is um negative two r sine theta and the k hath component is just this which is r quos and squared plus r sine squared which is r and now we can call this the area.

02:35 We're going to find the magnitude of this, the magnitude of r times r -sophdata to equal to the square root of 4 r squared, cossate squared data plus sine squared data, factored out, right, plus r squared.

02:51 Pretty much we can factor it out the square of r squared which is r, and this becomes r times the square root of 4 plus 1 is 5 square.

03:01 5 here...

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