Surface integral ex3 part 2: Evaluating the outside surface | Multivariable Calculus | Khan Academy

Khan Academy
29 May 201209:10
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TLDRIn the video, the presenter delves into the process of evaluating a surface integral, focusing on 'surface two' of a geometric figure. They detail a parametrization method, simplify the integral to du dv, and then proceed to compute the double integral over u and v. By leveraging trigonometric identities and carefully handling the bounds of integration, the presenter successfully calculates the surface integral, resulting in a final value of 3 pi over 2. The video concludes with a teaser for the next topic, which will cover 'surface three'.

Takeaways
  • ๐Ÿ“Œ The discussion continued from the previous video, focusing on evaluating a surface integral over 'surface two'.
  • ๐Ÿ”ต A parametrization for the blue side of a chopped cylinder was established in the previous session.
  • ๐Ÿ“ˆ The surface element dS was simplified to du dv for this particular surface.
  • ๐Ÿงฉ The surface integral was expressed as a double integral over u and v parameters.
  • ๐ŸŽจ Two different colors were used to distinguish between the u and v integration variables.
  • ๐ŸŒŸ The parametrization led to z being equivalent to v, simplifying the integral to an expression of v.
  • ๐Ÿ”„ The order of integration was chosen to be dv first, then du, based on the bounds of the parameters.
  • ๐Ÿ“Š The bounds for v were determined to be between 0 and 1 minus cosine of u, which is a function of u.
  • ๐ŸŒ€ The bounds for u were from 0 to 2 pi, simplifying the integration process.
  • ๐Ÿงฎ The antiderivative of v was calculated as v squared over 2 and evaluated from 0 to 1 minus cosine of u.
  • ๐Ÿฅ‡ The final result of the surface integral for surface two was found to be 3 pi over 2.
  • ๐Ÿš€ The next steps involve tackling the evaluation of 'surface three' in the subsequent video.
Q & A

  • What is the main topic of the video script?

    -The main topic is calculating the surface integral over a specific surface (surface two) of a three-dimensional object, described as a 'chopped cylinder', using parametrization.

  • How is the parametrization of surface two represented in the script?

    -The parametrization of surface two is simplified to dS = du dv, indicating a straightforward relationship between the surface element dS and the parameters u and v.

  • What integral expression is being evaluated in the script?

    -The script evaluates the surface integral of z dS over surface two, where z is parametrized as a function of v, leading to a double integral in terms of u and v.

  • Why was v chosen for the inner integral during the evaluation?

    -The choice to integrate with respect to v first was based on the bounds of v, which are from 0 to 1 - cos(u), indicating that v's upper limit is a function of u, making the integration process more straightforward.

  • What are the bounds for the parameters u and v in the integral?

    -For v, the bounds are from 0 to 1 - cos(u). For u, the bounds are from 0 to 2ฯ€.

  • How is the integral of z dS, with z = v, simplified before integration?

    -Before integration, the integral of z dS, with z = v, is simplified to v^2/2 evaluated from 0 to 1 - cos(u), and further simplified using algebraic manipulation.

  • What trigonometric identity is used to simplify the integral of cos^2(u)?

    -The trigonometric identity used is cos^2(u) = 1/2 + 1/2 cos(2u), facilitating the integration of cos^2(u) by breaking it down into more manageable terms.

  • What is the final result of the surface integral over surface two?

    -The final result of the surface integral over surface two is 3ฯ€/2, after evaluating all the integrals and simplifying.

  • Why does the integral of sin(u) from 0 to 2ฯ€ evaluate to 0?

    -The integral of sin(u) from 0 to 2ฯ€ evaluates to 0 because the sine function is symmetrical over this interval, with equal areas above and below the x-axis canceling each other out.

  • What is the significance of the computed surface integral value?

    -The computed value of 3ฯ€/2 for the surface integral represents a quantitative measure of the specified surface (surface two) within the context of the three-dimensional object being analyzed, potentially contributing to a larger calculation involving multiple surfaces.

Outlines
00:00
๐Ÿ“š Surface Integral Calculation for Chopped Cylinder

The paragraph begins with a recap of the previous video where the focus was on calculating the surface integral for 'surface two' of a chopped cylinder. A parametrization was found, leading to the simplification of dS to du dv. The surface integral is then expressed as a double integral over u and v, with z equated to v in the parametrization. The integration strategy is outlined, with v integrated first due to its variable upper bound dependent on u. The boundaries for v are determined as 0 and 1 minus cosine of u, with u ranging from 0 to 2 pi. The integral is simplified to a double integral, and the process of evaluating the integral is described, leading to the conclusion that the surface integral for surface two is 3 pi over 2.

05:01
๐Ÿ“ Detailed Evaluation of Surface Integral Components

This paragraph delves into the detailed evaluation of the surface integral components. It starts by rewriting the integral in terms of u and v, applying a trigonometric identity to transform the term involving cosine squared u into a more manageable form involving cosine of 2u. The antiderivative of this expression is then taken, which simplifies to u evaluated from 0 to 2 pi, resulting in 2 pi. The other components of the integral, involving cosine of u and cosine squared u, are evaluated using standard antiderivative techniques and trigonometric identities. The final evaluation results in the integral's value being 2 pi, and the paragraph concludes with the anticipation of tackling the next surface in the subsequent video.

Mindmap
Keywords
๐Ÿ’กSurface integral
A surface integral is a mathematical concept used to calculate quantities over a surface, rather than a volume. In the context of the video, the surface integral is the main focus as the presenter is working through evaluating it for a specific geometric shape. The process involves parametrizing the surface and then setting up a double integral to find the value of the integral.
๐Ÿ’กParametrization
Parametrization is the process of representing a mathematical object, such as a curve or surface, using a set of parameters. In the video, the presenter has found a parametrization for the surface they are working with, which allows them to express the surface in terms of variables u and v, simplifying the calculations.
๐Ÿ’กdS
dS represents an infinitesimal element of surface area, which is a fundamental concept when calculating surface integrals. In the video, the simplification of dS to du dv is crucial for setting up the double integral that will be used to evaluate the surface integral.
๐Ÿ’กDouble integral
A double integral is a mathematical operation that calculates the integral of a function over a two-dimensional region. In the video, the surface integral is expressed as a double integral over u and v, which are the parameters used in the parametrization of the surface.
๐Ÿ’กBounds
Bounds refer to the limits or intervals over which a variable is integrated in a double integral. In the video, the presenter discusses the bounds for the integration with respect to v, which depend on the function of u, and the bounds for u, which are from 0 to 2 pi.
๐Ÿ’กAntiderivative
An antiderivative, also known as an indefinite integral, is a function that represents the reverse process of differentiation. In the context of the video, the presenter is looking for the antiderivative of various functions to evaluate the double integral.
๐Ÿ’กTrigonometric identities
Trigonometric identities are equations that relate different trigonometric functions. In the video, the presenter uses a trigonometric identity to transform the term 'cosine of squared u' into a more manageable form, '1/2 plus 1/2 cosine of 2u', to simplify the evaluation of the integral.
๐Ÿ’กChain rule
The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. It states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In the video, the chain rule is implied when the presenter discusses taking the derivative of sine of 2u.
๐Ÿ’กEvaluation
Evaluation in the context of this video refers to the process of calculating the value of an expression or function at specific points. The presenter evaluates the antiderivative of various functions at the bounds of the integration to find the value of the surface integral.
๐Ÿ’กSine and Cosine
Sine and cosine are basic trigonometric functions that relate the angles and sides of a right triangle. In the video, these functions are used in the evaluation of the double integral, particularly in the context of evaluating the antiderivative of 'cosine squared u' and 'sine of 2u'.
๐Ÿ’กPi
Pi (ฯ€) is a mathematical constant approximately equal to 3.14159, representing the ratio of a circle's circumference to its diameter. In the video, pi appears as a result of evaluating certain integrals and is part of the final answer for the surface integral.
Highlights

The video focuses on evaluating a surface integral over a parametrized surface, specifically surface two.

A parametrization for the surface was previously found, simplifying the expression for dS to du dv.

The surface integral is expressed as a double integral over u and v, with z being equal to v in the parametrization.

The choice to integrate with respect to dv first is based on the bounds of the parameters, with v having variable upper bounds depending on u.

The boundaries for the integration with respect to v are 0 and 1 minus cosine of u, which is a function of u.

The integration with respect to u has the bounds of 0 to 2 pi.

The antiderivative of v is v squared over 2, and the integral is evaluated from 0 to 1 minus cosine of u.

The double integral is simplified to a straightforward computation, with the outside integral being from 0 to 2 pi of du.

The inner integral's antiderivative is v squared over 2, evaluated from 0 to 1 minus cosine of u.

The final expression for the surface integral is simplified to 1/2 times the difference of cosine u squared and 2 times cosine of u.

The integral evaluates to 1/2 times 2 pi minus 0 plus pi, resulting in a final value of 3 pi over 2 for the surface integral of surface two.

The video demonstrates a clear and methodical approach to evaluating surface integrals in a mathematically rigorous way.

The use of trigonometric identities, such as the one converting cosine squared to cosine of 2u, is highlighted in simplifying the integral.

The video provides a step-by-step walkthrough of the integration process, making it accessible for viewers to follow along.

The importance of correctly identifying and applying the bounds of integration is emphasized.

The video showcases the practical application of mathematical concepts such as parametrization and surface integrals.

The problem-solving approach in the video is systematic, breaking down complex integrals into manageable parts.

The video ends with a preview of the next topic, which will involve tackling surface three.

Transcripts

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MathematicsSurface IntegralParametrizationDouble IntegralCalculusGeometryCylinderEducationalProblem SolvingMath Tutorial