B)Determine the volume of the solid obtained by rotating the region bounded by :y=x -2x and y=xabout the line y- 4SolutionFirst let's get the bounding region and the solid graphed.Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about ahorizontal axis we know that the cross-sectional area will be a function of x.Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketchon the left includes the back portion of the object to give a little context to the figure on the right.Now, we 're going to have to be careful here in determining the inner and outer radius as they aren't going to bequite as simple they were in the previous two examples.Let's start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distancefrom the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis ofrotation to the inner edge of the ring So, we know that the distance from the axis of rotation to the x-axis is 4and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference betweenthese two.1) inner radius2) The outer radius is:3) The cross-sectional aren: A = a [f - g² ]outer´innerradius)radius,Using the formulas:Volume: "-S,a(x)dx4) Find the volume of this solid: , Discs and washers: Volume of solidsDetermine the volume of the solid obtained by rotating the region bounded by y=x² -4x+5X= 1, x= 4 and the x and y axes:Step 1:The first thing to do is get a sketch of the bounding region and the solid obtnined by rotating the region about the x-axis. Here are both of these sketches.Okay, to get a cross section we cut the solid at any x. Below are a couple of sketches showing a typical cross section. Thesketch on the right shows a cut away of the object with a typical cross section without the caps. The sketch on the leftshows just the curve we're rotating as well as its mirror image along the bottom of the solid.In this case the radius is simply the distance from the x-axis to the curve and this is nothing more than thefunction value at that particular x as shown above. The cross-sectional area is then,A(x) = #(x°- 4x+ 5)° = 7 (x* - 8x² + 26x² - 40x +25)a) Find the volume of the solid above: V=S a(x) dx

Question

B)Determine the volume of the solid obtained by rotating the region bounded by :y=x -2x and y=xabout the line y- 4SolutionFirst let's get the bounding region and the solid graphed.Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about ahorizontal axis we know that the cross-sectional area will be a function of x.Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketchon the left includes the back portion of the object to give a little context to the figure on the right.Now, we 're going to have to be careful here in determining the inner and outer radius as they aren't going to bequite as simple they were in the previous two examples.Let's start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distancefrom the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis ofrotation to the inner edge of the ring So, we know that the distance from the axis of rotation to the x-axis is 4and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference betweenthese two.1) inner radius2) The outer radius is:3) The cross-sectional aren: A = a [f - g² ]outer´innerradius)radius,Using the formulas:Volume: &#34;-S,a(x)dx4) Find the volume of this solid: , Discs and washers: Volume of solidsDetermine the volume of the solid obtained by rotating the region bounded by y=x² -4x+5X= 1, x= 4 and the x and y axes:Step 1:The first thing to do is get a sketch of the bounding region and the solid obtnined by rotating the region about the x-axis. Here are both of these sketches.Okay, to get a cross section we cut the solid at any x. Below are a couple of sketches showing a typical cross section. Thesketch on the right shows a cut away of the object with a typical cross section without the caps. The sketch on the leftshows just the curve we're rotating as well as its mirror image along the bottom of the solid.In this case the radius is simply the distance from the x-axis to the curve and this is nothing more than thefunction value at that particular x as shown above. The cross-sectional area is then,A(x) = #(x°- 4x+ 5)° = 7 (x* - 8x² + 26x² - 40x +25)a) Find the volume of the solid above: V=S a(x) dx

Accepted Answer

Answered: Image /qna-images/answer/1c383857-060b-44d8-aa82-fb09a5ab6e00.jpg