Double integrals 3 | Double and triple integrals | Multivariable Calculus | Khan Academy
TLDRIn this mathematical exploration, the video delves into the process of calculating the volume between the surface defined by xy squared and the xy-plane within specified limits for x and y. Initially, integration with respect to x was performed, resulting in a volume of 2/3. The video then demonstrates an alternative approach by integrating with respect to y first, holding x constant, and confirming the same result. This methodological validation reinforces the concept of double integration in determining volumes and showcases the power of mathematical consistency.
Takeaways
- π The problem involves calculating the volume between the surface defined by the equation z = xy^2 and the xy-plane over the region 0 β€ x β€ 2 and 0 β€ y β€ 1.
- π The method used in the previous video was integrating with respect to x first, then y, but this time we will reverse the order of integration to verify the result.
- π¨ The visualization of the problem involves a 3D graph with the x-axis, y-axis, and z-axis, where the xy-plane is at the bottom and the surface lies above it.
- ποΈ The first step in the new method is to hold x constant and calculate the area under the curve for a given x value, treating x as a constant.
- 𧩠The area under the curve for a fixed x is found by integrating dy * (xy^2) from y=0 to y=1, which gives us the area as a function of x.
- π The antiderivative of y^2 with respect to y is (y^3)/3, and when evaluated from y=0 to y=1, it results in the function (x/3) for a given x.
- π’ To find the volume, the area function (x/3) is multiplied by dx, and then integrated with respect to x from 0 to 2.
- π The antiderivative of x with respect to x is (x^2)/2, and when evaluated from x=0 to x=2, it results in the volume formula (x^2)/6.
- π The final volume calculated by integrating with respect to y first and then x is 4/6, which simplifies to 2/3, matching the result from the previous method.
- π This demonstrates the principle of mathematical consistency, where the same result can be obtained by differentεζ³η integration orders.
- π The video emphasizes the importance of understanding the order of integration and the ability to visualize mathematical problems in three dimensions.
Q & A
What was the main topic of the video?
-The main topic of the video was calculating the volume between the surface defined by the equation z = xy^2 and the xy-plane over the region where x ranges from 0 to 2 and y ranges from 0 to 1.
How did the video demonstrate the concept of integrating with respect to x first?
-The video demonstrated the concept by first fixing a value of y, then calculating the area under the curve for that y, and finally integrating with respect to x to find the volume.
What was the initial method used to calculate the volume?
-The initial method used to calculate the volume was by integrating with respect to x first and then integrating with respect to y.
Why was it important to try integrating in a different order?
-Integrating in a different order was important to verify the correctness of the result, showing that the volume calculated would be the same regardless of the order of integration.
How did the video explain the process of integrating with respect to y first?
-The video explained the process by holding x constant, treating it as a constant, and then calculating the area under the curve for a given x. This area was then multiplied by dx to account for the depth, and integrated with respect to x to find the volume.
What was the antiderivative of y^2 when integrating with respect to y first?
-When integrating with respect to y first, the antiderivative of y^2 was y^3/3, considering x as a constant.
How did the video represent the change in area depending on the value of x?
-The video represented the change in area depending on the value of x by showing that the area is x/3 for any given x, and this area changes as x varies from 0 to 2.
What was the final result of the volume calculation?
-The final result of the volume calculation was 2/3, which was the same as when integrating with respect to x first and then y.
How did the video conclude the demonstration?
-The video concluded by showing that the volume under the surface z = xy^2 and the xy-plane is 2/3, confirming that the result was consistent regardless of the order of integration.
What was the significance of the graph and surface visualization in the video?
-The graph and surface visualization in the video helped to provide an intuitive understanding of the volume being calculated and how the integration process relates to the shape of the surface.
What was the antiderivative of x with respect to the final volume calculation?
-In the final volume calculation, the antiderivative of x was x^2/2, which was used to evaluate the area of the green surface depending on the value of x.
Outlines
π Double Integral for Volume Calculation
This paragraph discusses the process of calculating the volume between a surface defined by the equation xy squared and the xy-plane with given constraints on the x and y axes. Initially, the speaker explains the method previously used, which involved integrating with respect to x first and then y, resulting in the answer 2/3. The speaker then introduces an alternative approach by integrating with respect to y first, to verify the same result. The explanation includes a detailed description of the graph and the steps to calculate the area under the curve for a fixed value of x, followed by integrating with respect to x to find the volume. The paragraph concludes with a reiteration of the graph and the calculated volume, emphasizing the importance of understanding the order of integration in double integrals.
π Evaluating Inner and Outer Integrals for Volume
The second paragraph delves into the specifics of evaluating the inner integral with respect to y, holding x constant, and then using the result to calculate the total volume. The speaker clarifies that the area represented in green is a function of x and depends on the chosen value of x. The process of evaluating the inner integral is explained, leading to the antiderivative of y squared, which is y cubed over 3 when x is constant. The outer integral with respect to x is then discussed, with the antiderivative of x being x squared over 2. The final step involves evaluating this expression at the bounds of integration (from 0 to 2) and multiplying by 1/3 to get the volume, which is confirmed to be 2/3, consistent with the previous method. The paragraph ends with a reflection on the successful completion of the volume calculation and a visual appreciation of the graph.
Mindmap
Keywords
π‘Volume
π‘Integration
π‘xy-plane
π‘Surface
π‘Antiderivative
π‘Limits of Integration
π‘Curve
π‘Area
π‘Depth
π‘Graph
π‘Coordinate System
Highlights
The video begins with a recap of a previous problem involving the volume between a surface and the xy-plane, specifically when x ranges from 0 to 2 and y from 0 to 1.
The method previously used involved integrating with respect to x first and then with respect to y, resulting in a volume of 2/3.
The video aims to confirm the correctness of the previous result by integrating in the opposite order, starting with respect to y.
The graph of the function is redrawn to provide visual context, emphasizing the x-axis, y-axis, and z-axis.
The xy-plane is defined with y ranging from 0 to 1 and x from 0 to 2, and the surface graph is approximated.
The focus is on the volume underneath the graph, which is the area under the curve integrated over the range of x.
A new method is introduced by holding x constant and treating it as a function of y squared, simplifying the area calculation.
The area under the curve for a given x is calculated by multiplying xy squared by dy and integrating from y=0 to y=1.
The volume under the entire surface is found by multiplying the area by dx, effectively adding depth to the 2D area.
The inner integral with respect to y is evaluated, resulting in a function of x, which is x/3.
The outer integral with respect to x is then calculated, with the antiderivative of x being x squared over 2.
The final volume calculation is performed by evaluating the x integral from 0 to 2, resulting in 4/6 or 2/3.
The video concludes by confirming that integrating in either order yields the same result, demonstrating the consistency of the mathematical approach.
The video ends with a visual representation of the rotated graph, reinforcing the understanding of the volume between the surface and the xy-plane.
The video serves as a practical example of applying integration techniques to find volumes, showcasing the importance of understanding different approaches.
Transcripts
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