# Homework Help: Solids of revolution

1. Feb 8, 2009

### nameVoid

y=x^2 ;
y=4;
rotated around x=2

im seeing a washer cross section with r=2-y^(1/2);
im unclear on how to get R to calculate the area it seems to be 2r but this produces incorrect results.

2. Feb 8, 2009

### hokie1

y=x^2 is a parabola and y=4 is a line. Rotate these about the x=2 axis and the object is a paraboloid.

3. Feb 8, 2009

### nameVoid

im clear on the shape of the solid , but how do you get an equation for the large radius of the washer

4. Feb 8, 2009

### Staff: Mentor

When you rotate the region of the parabola y = x^2 around the line x = 2, the vertical cross-section of this solid looks like two parabolic sections. The part on the right has its vertex at (4, 0) and intersects the line y = 4 at (2, 4) and (6, 4). The equation of this translated parabola is y = (x - 4)^2.

The large radius of a washer is the x-value on the parabola on the right, minus 2, or sqrt(y) + 4 - 2. You can also get this dimension by taking the x value on the parabola on the left, and you'll get the same value.

You mentioned that you had calculated the large radius as 2 - sqrt(y). That actually gets you the inner radius. Using the parabolic region on the right, I get an inner radius of 4 - sqrt(y) - 2, which is what you had for the outer radius.