The washer method uses an adaptive version of the volume of a cylinder.  If we think of a washer, it is disc-like but has a hole in the middle.  So we need to consider both an inner and outer radius.  Similar to the disc method we are taking cross sectionals of a solid but this solid has a hole in the middle.  We sum infinitely many of these cross sections to obtain a value for the volume.

The washer method uses an adaptive version of the volume of a cylinder.  If we think of a washer, it is disc-like but has a hole in the middle.  So we need to consider both an inner and outer radius.  For this reason, you must consider two independent functions as two separate radii measured as the distance from the axis of rotation.

When trying to think about our integration limits remember that  this function starts at the origin and moves in the positive  and  direction, so our lower limit will be .  When thinking about our upper limit we must find when these two functions intersect.  To do so, we set them equal to one another:

So our upper limit is .  Next we must think about the volume of the solid we are trying to find.  We will have two radii to consider since there are two different  equations.  Let’s rearrange both in terms of .

The function that would be ‘on top’ in a graph would be , so this is our inner radius ( ) making our outer radius ( ).

Now we can plug into our washer method formula 

Explanation: First we must try to find our integration limits.  So we need to find where these two functions intersect.  We do so by setting them equal to one another.

So  is one limit.  If we look at both functions we see they both pass through the origin, so  is our lower limit and  is our upper limit.  Next we must think about the volume of the solid we are trying to find.  We will have two radii to consider since there are two different  equations.  Let’s rearrange both in terms of .

The function that would be ‘on top’ in a graph would be .  So this is our inner radius ( ) making  our outer radius( ).

Now we can plug into our washer method formula 

Now we are considering a solid who is not rotating around the x or y-axis but the line .  First we must find the limits of integration.  We can look at their points of intersection to find the upper and lower limits.  We do this by setting them equal to one another.  

So our lower limit is  and the upper limit is .

Now we must determine our radii.  Our inner radius ( ) will be :

For the outer radius ( ):

Now we can plug this into our washer method formula  

Because the disk is of radius R, the base is defined by the following formula: . 

The correct formula for the area of an equilateral triangle is as follows:

, with  being the side length of the triangle.

By applying this formula to our general volume formula , we get the following: .

The radius R defines the bounds as being . Next, s can be found by understanding that the value is the distance from the top to the bottom of the circle at any given point along . The length of one side of the equilateral triangle, therefore, is .

Putting it all together, the following is obtained:

*Note: the problem did not specify if the cross sections were perpendicular to the  or axis. Because the base is a circle, this should not change the resulting volume. The only difference should be the use of  or as variables in the correct expression.

Because the base is a circle of radius , the bounds are defined as .

The area of a right isosceles triangle can be found using the formula , where  is the leg length of the triangle. By applying this to our general volume formula , we get the following: . 

The expression for  can be found by understanding the fact that the leg  of the triangle is on the base of the solid. The value is twice the height of the semicircle . 

Putting it all together, the following is obtained:

First, the cross sections being perpendicular to the  axis indicates the expression should be in terms of . 

The area of an equilateral triangle is , with  being the side length of the triangle. By applying this formula to our general volume formula (), we get the following: .

The intersection points of the functions  and  are  and . The  coordinates of these points will define the bounds for the integral, since our expression is in terms of .

The base is bounded by  and . Rewriting these functions in terms of , the following equations are obtained:  and . Since  is farther from the  axis, the correct expression for the side length is .

Putting it all together, the following is obtained:

First, the cross sections being perpendicular to the  axis indicates the expression should be in terms of . The area of a right isosceles triangle can be found using the formula , where  is the leg length of the triangle. By applying this to our general volume formula , we get the following: .

The intersection points of the functions defining the region are  and . The  coordinates of these points will define the bounds for the integral, since our expression is in terms of .

The base is bounded by ,  and . Since the cross-sections are perpendicular to the  axis, the leg of the triangle cross-sections are defined by: .

Putting it all together, the following is obtained:

Because the disk is of radius , the base is defined by the following formula: . 

The correct formula for the area of an equilateral triangle is as follows:

, with s being the side length of the triangle.

By applying this formula to our general volume formula , we get the following: .

The radius  defines the bounds as being . Next,  can be found by understanding that the value is the distance from the top to the bottom of the circle at any given point along . The length of one side of the equilateral triangle, therefore, is .

Putting it all together, the following is obtained:

*Note: the problem did not specify if the cross sections were perpendicular to the  or axis. Because the base is a circle, this should not change the resulting volume. The only difference should be the use of   or  as variables in the correct expression.