Volume+with+known+cross+sections

=Volume with Cross Sections= We have seen how the Volume of a solid of revolution can be calculated using integration. Now we will find the Volume of a new type of shape using integration: Solids whose cross sections are always a particular shape.

In the previous section, you noticed that we summed the areas of the circles created by the revolution. Now the area formula will change depending on the cross-section's shape.

In these problems, functions that define the regions to be integrated will give us the value of needed dimensions of the cross section to find the cross-section's area. This is called the Special Length. For example, our function may define the diameter when the cross sections are semi-circles, or the side length of a triangle.

The solid is bounded by planes perpendicular to the x-axis at x=0 and x=5 and the curves y=(-sqrt(x)) and y=sqrt(x). Find the Volume of the solid created whose cross sections are (All cross sections are perpendicular to the x-axis)
 * Example One **
 * A**. squares whose base is in the xy plane, **B**. Equilateral triangles whose base is in the xy plane, **C**. Semicircles whose diameter is in the xy plane

Solution: The area formula for a square is A=s 2, so we can now sum the areas of squares for the given interval 0<x<5. Now we could integrate the areas over the given interval. However, notice that the only difference in this formula from the square formula is the sqrt(3)/4 constant. Therefore, we can multiply part A's answer by this constant (as the constant would move out front of the integral), yielding **V=25sqrt(3)/2 u 3 **.
 * A**. The base extends between the two curves, so we have a side length of sqrt(x)-(-sqrt(x))=2sqrt(x)=sqrt(4x).
 * B**. The side length is the same as part A, but the Area formula for Equilateral triangles is A=s 2 x sqrt(3)/4.

Note that the only difference between this formula and the original formula is the p /8 constant, so we will multiply part A's answer by p /8, yielding **V=25p /2** **u 3 **.
 * C**. Now the area formula is (1/2)p r 2 =(1/2)p (d/2) 2 =(p /8)d 2, and our Special Length is the diameter of the semicircle.