mass of the lamina

Question

F‎‎‎‎‎‎‎‎ind the mass of the lamina in the shape of the portion of the plane with equation 4x + 8y + z = 8 in the ‎‎‎‎‎first octant if the area density at any point (x, y, z) on the plane is δ(x, y, z) = 6x + 12y + z g/cm^2.

Answer 1

We need to integrate the density function δ(x,y,z) over the portion of the plane in the first octant.

That portion forms a triangle; first calculate some dimensions of that triangle.

Let X be the point where the plane intersects the x-axis, i.e., where y=z=0:

4x + 8×0 + 0 = 8 ==> x = 2

X = (2, 0, 0)

Let Y be the point where the plane intersects the y-axis, i.e., where x=z=0:

4×0 + 8y + 0 = 8 ==> y = 1

Y = (0, 1, 0)

Let Z be the point where the plane intersects the z-axis, i.e., where x=y=0:

4×0 + 8×0 + z = 8 ==> z = 8

Z = (0, 0, 8)

Let O be the origin

O = (0, 0, 0)

From the triangle OXY use Pythagorean theorem to calculate the length of XY:

XY = √(OX² + OY²) = √(2² + 1²) = √5

From the point O drop a perpendicular to XY.

Let P be the point where the perpendicular meets XY.

We can calculate the length OP because there are two ways of calculating the area

of the triangle OXY:

OX×OY/2 = XY×OP/2

OP = OX×OY/XY = 2/√5

From the triangle OZP use Pythagorean theorem to calculate the length of ZP:

ZP = √(OZ² + OP²) = √(8² + (2/√5)²) = 18/√5

Having calculated all we need regarding the triangle XYZ, we can now calculate its mass.

To do the integration imagine the given plane divided into horizontal stripes -- one

for each value of z between 0 and 2.

Each stripe has width of some diminishingly small amount ds.

Let dz be the projection of ds on the z-axis.

Since ds is along the plane, at an angle with respect to the z-axis, dz < ds.

By similarity of triangles, ds is related to dz same as ZP is related to OZ.

Therefore ds = (ZP/OZ)dz = ((18/√5)/8)dz = (9/4√5)dz.

Divide each stripe into small rectangles, each with diminishingly small length dh.

Let dy be the projection of dh on the y-axis.

By similarity of triangles, dh is related to dy same as XY is related to OY.

Therefore dh = (XY/OY)dy = (√5/1)dy = √5dy

Consider one of these rectangles at a position (x, y, z).

Its area is dsdh and the density at that point is δ(x,y,z), so the mass of the rectangle is

m(x,y,z) = δ(x,y,z)dsdz = δ(x,y,z)(9/4√5)dz√5dy = δ(x,y,z)(9/4)dydz

Since the point (x,y,z) lies on the plane, we can replace

x = (8 - 8y - z)/4 = 2 - 2y - z/4

Substituting that into the definition of δ(x,y,z) we get

m(x,y,z) = (6(2 - 2y - z/4) + 12y + z)(9/4)dydz = (27 - 9z/8)dydz

The mass if the lamina is then the definite integral, where z ranges from 0 to 8, and y ranges from 0 to 1.

∫∫(27 - 9z/8)dydz =

∫(27 - 9z/8)(∫dy)dz =

∫(27 - 9z/8)dz =

(27z - 9z²/16) between 0 and 8 = 180 g

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The mass of the lamina is then the definite integral, where z ranges from 0 to 8.

For each value of z, the variable y ranges from 0 to where the stripe at height z

intersects the line ZY. That is, where y satisfies

4×0 + 8y + z = 8

So below, the bounds of the first integral (over z) are 0 to 8 and the bounds of the

second integral (over y) are 0 to 1-z/8.

∫∫(27 - 9z/8)dydz =

∫(27 - 9z/8)(∫dy)dz =

∫(27 - 9z/8)(1-z/8)dz =

∫(27 - 9z/2 + 9z²/64)dz =

(27z - 9z²/4 + 3z³/64) between 0 and 8 = 96 g