For instance, if someone could tell me how to even set up the integral for the volume of the rotation of region defined by the curves 27y=x3, y = 0, and x = 3 rotated about y = 1 (using the cylindrical shell method), you will have officially helped me more than my professor has in the past week.
The assignment is due in 15 minutes and I have no idea where to begin asking for help on the 4 remaining problems. I'll be fine losing the points, but that doesn't change the fact that the material was not taught thoroughly enough. My lecture notes are primarily filled with Riemann sums that demonstratewhywe can use an integral to find a solid of rotation, but our professor never has time to show us HOW to do so...which is what we need to know.
For instance, if someone could tell me how to even set up the integral for the volume of the rotation of region defined by the curves 27y=x3, y = 0, and x = 3 rotated about y = 1 (using the cylindrical shell method), you will have officially helped me more than my professor has in the past week.
FFFFFFFF do you have Skype? just call me or something I'll explain it as fast as I can.
For instance, if someone could tell me how to even set up the integral for the volume of the rotation of region defined by the curves 27y=x3, y = 0, and x = 3 rotated about y = 1 (using the cylindrical shell method), you will have officially helped me more than my professor has in the past week.
The obvious first step is to draw out the area you're rotating, which we can assume you've done.
However, we're rotating about y = 1, so that means that we want x in terms of y for integration. Furthermore, the basic cylindrical shell method is easiest to apply about an axis, so it would be sensible to shift our co-ordinates over so that we're rotating about y' = 0 (Note: y' is a shifted y, not to be confused with the derivative of y). The area of interest is between y = 0 and y = 1, so we will chose y' = 1 - y as our new co-ordinate. With our new co-ordinates, the region of interest is defined by the curves x = f(y') = 3(1-y')1/3, y' = 1, and x = 3, rotated about y' = 0. Hopefully this will be enough for you to apply the method.
EDIT: Pity it's too late, but you ought to solve the problem anyway.
Hopefully this will be enough for you to apply the method.
EDIT: Pity it's too late, but you ought to solve the problem anyway.
Yeah, it's at the very least aiding my understanding. In terms of shifting the axis of rotation, I take it that if one wished to rotate about y = -1, my y' = y+1?
Yeah, that would work. I could also have chosen y' = y - 1 for your problem, but y' = 1 - y was nicer since it meant the region of integration was bounded from y = 0 to y = 1.
The next step is to express the volume as a Riemann sum of cylindrical shell elements (This simplifies down to a common integral formula, of course).
Also, you're probably going to want to go back to using the non-shifted y when you actually do the integral, since (1-y')1/3 is not the most convenient of things to integrate, but shifting the co-ordinates in that manner ought to help with understanding.
*sigh* I haven't done this kind of calc in almost 4 years and it shows. Just tried to do it on a post-it and it feels like I'm not setting it up right... oh well.
I'm still perplexed on how to change the axis of rotation...I'm waiting on an email from my TA to explain it in more depth. I've never actually witnessed the notation y' nor the idea of a "shifted y" before. Usually in class such shifts of the axis are just expressed as defining the height of the cylinder as x +/- c, where c is the axis we're trying to rotate around. I just don't exactly know when to add or subtract c (yes, I know that likely sounds dumb, and I know I'll feel stupid when I figure it out).
It might help to look at what the volume element is for the cylindrical shell method first (the volume element being one of the things you're summing together).
Shifting the axis is just another way of looking at the same thing; if it's not how you've usually done it then that's fine too. In fact, it probably isn't a very normal way to do it since you'd just end up doing another substitution to do the actual integral, at least in this example. It was just the first way to do it that came to my mind.
Let me give an approximate expression: approximate volume of cylindrical shell, given radius r, height h, and thickness dr, is 2 * pi * r * h * dr
In the problem we're looking at, the only thing being changed by shifting the axis of rotation is the radius (r), so we could achieve the same effect by changing only this term in our integral. What is the radius? The distance between the axis of rotation and the edge of the cylindrical shell element.
Usually in class such shifts of the axis are just expressed as defining the height of the cylinder as x +/- c, where c is the axis we're trying to rotate around. I just don't exactly know when to add or subtract c (yes, I know that likely sounds dumb, and I know I'll feel stupid when I figure it out).
Shifting the axis of rotation doesn't change the height of the cylinder - the height is parallel to the axis of rotation and hence won't change. What it changes is the radius.
I'm still perplexed on how to change the axis of rotation...I'm waiting on an email from my TA to explain it in more depth. I've never actually witnessed the notation y' nor the idea of a "shifted y" before. Usually in class such shifts of the axis are just expressed as defining the height of the cylinder as x +/- c, where c is the axis we're trying to rotate around. I just don't exactly know when to add or subtract c (yes, I know that likely sounds dumb, and I know I'll feel stupid when I figure it out).
I've never been a fan of shifting axes myself, but you basically just say that y' is going to be my new plane. Let's say that I want to make the cylinder go around the 0-axis, so you're effectively shifting the whole plane up by one. So you're trying to relate the new plane y' to the old plane y. Their relationship is y' = 1 - y. The reason why I don't like doing this is because when you do a shift like this, you also have to change the limits, which if you do a shift down by one, the limits are now -1 to -2/3 instead of 0 to 1/3.
Personally I would prefer to just say that the circumference of the cylinder is 2(1-y)pi going from 0 to 1/3. That is effectively the same as saying 2(y')pi going from -1 to -2/3.
y' = 1 - y is a shift and a reflection, in fact, since we have a negative sign on the y. This might be too complicated an approach for some, so yeah, perhaps it's best to stick with simply changing the radius in the Riemann sum. Also, I think the limits are from 0 to 1, since 27y = x3.
y' = 1 - y is a shift and a reflection, in fact, since we have a negative sign on the y. This might be too complicated an approach for some, so yeah, perhaps it's best to stick with simply changing the radius in the Riemann sum. Also, I think the limits are from 0 to 1, since 27y = x3.
Ah, oops. I was doing x2 in my head for some reason. Yes, 0 to 1 is right.
EDIT: Yeah, 0 to 1 makes it much easier. Now I want to know what the answer is to see if I still got it.
If you are going to integrate over x, sure. But if you want to be able to use the cylinder method on an assignment, you need to be able to say what the equations represent.
If you are going to integrate over x, sure. But if you want to be able to use the cylinder method on an assignment, you need to be able to say what the equations represent.
Right, okay. So the same equation stands when integrating over y right? 2*pi*y*(f(y) - g(y))?
1. *Facepalm*. Sorry, I'm writing a paper at the same time on economic foreign relations between postwar Japan and the US. Sorry if I'm not all with it.
Forensic food microbiology strikes again! I have to work Saturday, Sunday, and both of the days that I had taken off next week (the day before and after T-day). Granted, it's at most an hour each day, but then again, it's an hour each day that I wasn't planning on working.
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However, we're rotating about y = 1, so that means that we want x in terms of y for integration. Furthermore, the basic cylindrical shell method is easiest to apply about an axis, so it would be sensible to shift our co-ordinates over so that we're rotating about y' = 0 (Note: y' is a shifted y, not to be confused with the derivative of y). The area of interest is between y = 0 and y = 1, so we will chose y' = 1 - y as our new co-ordinate. With our new co-ordinates, the region of interest is defined by the curves x = f(y') = 3(1-y')1/3, y' = 1, and x = 3, rotated about y' = 0. Hopefully this will be enough for you to apply the method.
EDIT: Pity it's too late, but you ought to solve the problem anyway.
The next step is to express the volume as a Riemann sum of cylindrical shell elements (This simplifies down to a common integral formula, of course).
Also, you're probably going to want to go back to using the non-shifted y when you actually do the integral, since (1-y')1/3 is not the most convenient of things to integrate, but shifting the co-ordinates in that manner ought to help with understanding.
Shifting the axis is just another way of looking at the same thing; if it's not how you've usually done it then that's fine too. In fact, it probably isn't a very normal way to do it since you'd just end up doing another substitution to do the actual integral, at least in this example. It was just the first way to do it that came to my mind.
Let me give an approximate expression:
approximate volume of cylindrical shell, given radius r, height h, and thickness dr, is
2 * pi * r * h * dr
In the problem we're looking at, the only thing being changed by shifting the axis of rotation is the radius (r), so we could achieve the same effect by changing only this term in our integral. What is the radius? The distance between the axis of rotation and the edge of the cylindrical shell element. Shifting the axis of rotation doesn't change the height of the cylinder - the height is parallel to the axis of rotation and hence won't change. What it changes is the radius.
Personally I would prefer to just say that the circumference of the cylinder is 2(1-y)pi going from 0 to 1/3. That is effectively the same as saying 2(y')pi going from -1 to -2/3.
Also, I think the limits are from 0 to 1, since 27y = x3.
EDIT: Yeah, 0 to 1 makes it much easier. Now I want to know what the answer is to see if I still got it.
The one for cylinders is 2*pi*(H-h)*r where H is the larger height and h is the smaller.