# A cement conduit 100 cm long is constructed so that it has a square exterior and a centered cylindrical interior. At its thinest point, the conduit's wall is 5 cm thick. Express the volume of...

A cement conduit 100 cm long is constructed so that it has a square exterior and a centered cylindrical interior. At its thinest point, the conduit's wall is 5 cm thick. Express the volume of cement needed to create such a conduit as a polynomial in terms of r, the radius of the conduit's interior.

steveschoen | Certified Educator

To do this, one must be aware of the formulas for the volume of a box, V = L*W*H, and a cylinder, V = pi * r^2 * H.  For, what we would effectively/mathematically "making the box" then "taking out the cylinder", aka subtracting the cylinder.

The visual I attached will probably be necessary to follow any explanation.  It shows a diagram of the conduit.  It is always a good idea to try to get a visual of the picture.

Since the cylinder is "centered", each red line is 5 cm long.  The radius is "r".  So, the distance to the side from the center of the hole is "r+5".  We can double that to find the length of the end, "2(r+5)".  And, since the end is a square, all sides of the end have a length of "2(r+5)".  So, the volume of the "box" would be:

V (box) = L*W*H = 2(r+5)*2(r+5)*100 = 400(r^2 + 10r + 25) = 400r^2 + 4000r + 10,000

Be weary of which is L, W, and H.  Formulas are always general.  Here, I made H the 100 cm distance.

Then, we need to subtract the interior cylinder.  The formula for that volume would be:

V (cylinder) = pi * r^2 * H = pi * r^2 * 100

Assuming pi = 3.14:

V (cylinder) = 314r^2

Subtracting the 2 formulas, we have:

(400r^2 + 4000r + 10,000) - (314r^2) = 86r^2 + 4000r + 10,000

I hope this helps.

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embizze | Certified Educator

The linear term should be 4000r (distributive property)