Khan Academy: Calculus: Surface Integral Ex1 Pt2: Calculate Surface Differential

Part 3 of a 3-part video series showing how to calculate the surface integral of the unit sphere. Video calculates the double integral of the magnitude of the cross product found in the previous part. Trigonometric identities and...

Second part of a two-part video series on calculating the surface integral of a surface that can be described as a function of x and y. Video evaluates the surface integral. [9:51]

Part 2 of a 4-part video series solving for the surface integral of a surface made of three separate parts. The second video finishes evaluating the surface integral of the second surface by substituting a trigonometric identity. [9:10]

First part of a two-part video series on calculating the surface integral of a surface that can be described as a function of x and y. Video shows parameterizing the surface. [5:10]

Part 1 of a 3-part video series showing how to calculate the surface integral of the unit sphere. The first part shows how to parameterize the unit sphere. [11:05]

Part 1 of a 4-part video series solving for the surface integral of a surface made of three separate parts. The first video sets up the surface integral for the entire surface, evaluates the first surface and parametrizes and finds the...

Part 3 of a 4-part video series solving for the surface integral of a surface made of three separate parts. The third video parametrizes the third surface, finds the magnitude of the cross product and begins to evaluate the double...

Part 4 of a 4-part video series solving for the surface integral of a surface made of three separate parts. This video finishes evaluating the surface integral of the third surface and adds the results together. [4:02]

Video lesson introducing double integrals by first reviewing definite integrals. Includes an understanding of how to use double integrals to calculate the volume under a surface using a visual representation but no mathematical...

Video lesson finishing an example of how to evaluate the double integral to find the volume under a specific surface when the bounds are variable. Initially integrates with respect to x first and then solves the problem again integrating...