The Disk Method Examples | Calculus 2 - JK Math
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- Äas pĹidĂĄn 16. 07. 2024
- Example Problems For How to Use The Disk Method To Calculate Volume (Calculus 2)
In this video we look at several practice problems of calculating the volume of a solid of revolution formed by revolving a region around the x-axis or y-axis by using definite integrals via the disk method.
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This series is designed to help students understand the concepts of Calculus 2 at a grounded level. No long, boring, and unnecessary explanations, just what you need to know at a reasonable and digestible pace, with the goal of each video being shorter than the average school lecture!
Calculus 2 requires a solid understanding of calculus 1, precalculus, and algebra concepts and techniques. This includes limits, differentiation, basic integration, factoring, equation manipulation, trigonometric functions, logarithms, graphing, and much more. If you are not familiar with these prerequisite topics, be sure to learn them first!
Video Chapters:
0:00 Example 1 - y=sqrt(9-x^2), y=0 around x-axis
8:04 Example 2 - y=2/x, y=1, y=8, x=0 around y-axis
15:27 Example 3 - y=e^x, x=0, y=0, x=ln(4) around x-axis
22:35 Example 4 - y=2x, y=3-x^2, x=0 around y-axis
33:40 Outro
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Understood now
Excellent Videos on disc and washer methods! helped me understand thanks!
You're welcome! Glad the videos were able to help you :)
Hii i would just like to ask, in the last example why was the two integrals added and not subtracted? don't we usually subtract two integrals?
In the last example we needed to break up the region being revolved around the y-axis into two smaller regions. We needed to do this because the outer function (or radius) of the region changes, it is not the same for the entire region. From y=0 to y=2, y=2x is the outer function, but then from y=2 to y=3, y=3-x² is the outer function. I explain this in more detail in the video, around the 27-28 minute mark. So, since we break up the region into two parts and we are calculating two different volumes that make up the solid of revolution, we need to add them together to get the volume of the entire solid of revolution.
I think what you are referring to with subtracting integrals is when you are calculating the area under a curve or when using the washer method. In each of those cases we are subtracting the area/volume formed by a lower function that we do not want to include, which is not what is happening in this video. Hope this helps!
In the last example, there are two regions \between the curves , yet we only solved for one. How come? Great vid btw!
Good question! One of boundaries for the enclosed region of interest that was given is x=0, which is the same as the y-axis. So when considering that line, as well as the other two curves, the region that is explicitly enclosed by them is region we work with in the video, as all three curves/lines make up some portion of the boundary of the region (and also remember we are told to look in the first quadrant only). While the region below y=2x and between the x-axis is also in the first quadrant, the x-axis is not mentioned as one of our boundaries (or y=0). So, considering the description in the problem, the region we want must be between the given curves and x=0, and be in the first quadrant, and the region that best fits that description is the one we work with. Hard to explain well with just words, but I hope this helps paints a clearer picture of why we worked with the region we did. Feel free to ask any further questions if this is still not clear.