Originally Posted by

**ChipB** Your equations for S3, S4 and S5 are incorrect. I'll help you with S3, then you can use the same technique to determine S4 and S5. Also, you should be aware that the values for S3 and S4 are not equal.

The equation of the circle that defines S3 comes from:

$\displaystyle (x-x_0)^2 + (y-y_0)^2 = R^2$

where (x_0,y_0) is the coordinates of the center of the circle and R is the radius. So for S3, which has a radius of 13 mm, we have:

$\displaystyle (x-25)^2 + (y-25)^2 = 13^2$

After some rearranging:

$\displaystyle y = \pm \sqrt{13^2-(x-25)^2} + 25$

Now we need to determine whether the sign in front of the square root is plus or minus. If we let x = 25, we get $\displaystyle y = \pm 13 + 25$ We're interested in only the lower value, to yield a value of y(25) = 12, so we know that the sign in front of the square root should be negative.

Finally, the height of the fillet is equal to h = y-12:

$\displaystyle h = -\sqrt {13^2-(x-25)^2} + 13$

The integral for S3 is then:

$\displaystyle S_3 = \int_{12} ^{25} ( -\sqrt {13^2-(x-25)^2} + 13) x dx$

Notice how the limits for the integral are not 0 to 13 as you had it - I am keeping a consistent definition of the x- and y- coordinates here.

Try this - see what you get for S3, then use the same technique for S4 and S5.