finding-the-volume-of-solids-of-revolution

Good day students welcome to the AP calculus AP exam review of the going to be going over problem, one of the first parts question four two thousand six and ASAP prowl addresses of cost of such as the area between curves of volume of so it's a revolution in volumes of solids with known cross-sectional areas so before we get started let's take a look at the of some formulas that are going to be guiding our problem-solving process. So start to it title it formulas right first one has to do with the area the area formula is basically the integral from eight to be of the function on the top is called F minus the function on the bottom is Koji to be the function to the right or the function to the left right how do we know the orientation will the X basically indicates the top down, orientation function in the top my is a function on the bottom in this case of the slice of the direct the cross-sectional area of the rectangle or region is going to be perpendicular to the X axis or DY the why the situation where you look at a function to the right minus a function to the left okay in that case your prospect your of rectangle or region will be pointing to the y-axis okay with this, orientation G is less than half okay unities to the bottom of F orgies to the left of F now volume is a little bit more of intricate stick a look at the formula for volume so volume of the integral from this pie times the integral from eight to be of the outer radius the are square minus the inner radius are square okay I gives us the area of the washer and then we integrate from a to be we can find appalling of the X is when you're of radius is perpendicular to the X axes or pointing to the X axis and the why is when you radius is perpendicular or points to the y-axis now there are two cases to scenarios here the painting on the orientation of your of axis of rotation so let's take a look at the two cases the first one pie times the integral from eight to be to be looking at a top-down orientation where top down the left five is the axes of rotation is less than the two functions are going to have of the function on the top or to the right minus the axes the rotation square this is the are square pie are square help you find the area of the outer of the bigger desk minus G minus a square this is the you are square to pie times that square will give the area of the inner desk in a we subtract these two areas you get the area of the washer okay so of the X if you radius is perpendicular to the X axes or DY if you're radius is perpendicular to the y-axis right so this situation is applicable when a is smaller than built functions either is in the bottom of G and F or is to the left of G and okay and then the other situation we have is pie times the integral from a to be of a minus F this is the art a different here is of a is above the bigger the both of the two functions case the a minus F square will give you the outer radius minus a minus T-square will give you the inner radius the X if you slice is perpendicular to the X axes a radius is perpendicular the X axis or DY if you slice is perfect because the y-axis is orientation is applicable when F is listing G NG is less than eight is like a reversal of the order here okay I would like to put F at and and into T in the middle and then mom a at the other end. So this formula is it applicable to of revolution solids of generated as a result of revolution now how about the volume of solids with known cross-sectional area that's of very straightforward that one. The volume is simply going to be the integral from a to be of the cross-sectional area cross-sectional area can be a have of you be of X or a of why independent so the X is the base is perpendicular to the X axis or DY is the base of the area is perpendicular to the y-axis so that's this is for him cross-sectional area is that that note the across section you we had part with this last of formula is finding the area of the cross-sectional area can you have to make use of your geometric formulas to find the area okay right let's take a look at problem, one a part two that are be to the shaded region bounded by the graph of why equals on X and the line why equals X minus two question about find the area of our okay so to find the area. The going list is good to have use the formulas to guide our problem-solving process. So that. The formula for area that we talked about earlier is the integral from eight to be of F the function on the top or to the right minus T the function on the bottom are to the left the X or DY now here we have a situation where regardless of how weak of credit a region of will going to have one function constantly on top and will function constantly on the bottom one function constantly to the left and will function constantly to the right so this

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