Volumes By Slicing
I have huge difficulties with integration (I'm decent with derivatives, thank god), and finding volumes and other applications of integration confuses the heck out of me. The prof really doesn't explain why things are the way they are, which I tend to need to fully understand how to do the question as well as questions like it.
The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross sections perpendicular to the x axis between these planes are squares whose bases run from the semi circle y= -sqrt(1-x^2) and y= sqrt(1-x^2). Find the volume.
I know that the functions given create the top and lower half of a circle, respectively. What I am really confused about with this question is how to draw the solid it asks for. I am told that the cross sections create squares... What do I do with this information?
The region enclosed by x= y^(3/2), x=0, y=2, rotated about the y axis.
So, I solve for the equation in terms of y, and I get y= x^(2/3) (first off, is this legal to do?). When I rotate this around the y axis, I get a cone-like shape. I know that the formula for finding the volume is the integral of pi times the radius squared, evaluated from y=0 to y=2, except that the radius is in terms of x. How should I go about solving this question?
A third one is:
Find the volume of the solid generate by revolving the regions bounded by the lines and curves about the x axis.
y=secx, y=tanx, x=0, x=1
I can't even picture this when it is made...
I can generally do the integration itself, I just have great difficulties developing the problem (including sketching certain functions, and how the data given to me fits into certain formulas).
Can anyone please help me work my way through this stuff? I really need to learn how to do it, and its just confusing the heck out of me... Thanks.
Essentially, when you study this stuff, you learn a new definition of volume. With complex shapes composed of curves, you can't simply compute height * width * length anymore. After all, they all change now and are no longer constant. So, we learn that in actuality, volume is merely the infinite summation of infinitesimally thin cross sections with certain areas. What we are doing is relating the areas of these cross sections with their distance along an axis. That is, as you move away from the origin on an axis, the area of the cross sections changes predictably.
Originally Posted by Hellreaver
In your first question, the cross sections are square. Now, the dimensions are delineated by the equation for a circle centered at the origin with a radius of one. Try and picture this in your mind's eye. Visualization is key to these problems. The bottom leg of the square is the height above the x-axis given by (1-x^2)^.5 and below the x-axis by -(1-x^2)^.5 This boundary is that of a circle and therefore symmetric about the x- axis. Therefore, the length of the bottom leg of the square cross-section has a length of 2(1-x^2)^.5 If that is the length of one side of the square, what is its area? Obviously, 4(1-x^2). Now, you have your area function. Simply integrate along the x-axis from -1 to 1 and there is your answer.
Second question. You can rewrite the equation sure. However, you'll need the equation in terms of the variable whose axis you will revolve about. If you are revolving about the y axis, then your equation needs to be in the form x= ..... not y=..... otherwise you'll end up with a mix of x and y in the integral and you can't have that. (Not yet anyway...wait until next semester!) So, keep it as x = y^(3/2) that's your integral. In these problems, you often need to hunt for the boundaries of integration. They are usually easy to find. Here you have x = 0 and y = 2. Well, we need everything in terms of y. So, when x = 0 what does y equal? 0 Square the integrand so as to construe it as a radius of an infinitesimal circular cross section and integrate (y^(3/2))^2 from 0 to 2. multiply by pi and bingo.
The third. Often, you will see this type of volume of revolution. The first two types discussed above are relatively simple. This third, however, is of a third, somewhat more sophisticated type. There are shapes that when revolved, produce cavities. This is one of those. If I am picturing this graph correctly in my mind's eye, this solid does not produce a completely solid shape. There is an opening like that of brass instrument. These volumes are calculated thus: square both functions as to construe each as the radius of the circular cross section of infinitesimal width. Take the squared "lower" function (the function lower on the graph grid) and subtract it from the squared "higher" function. This is your integral. You are calculating the volume created by the higher function minus that of the lower. Integrate and multiply by pi.