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Unformatted text preview: Math 23
Week 5a Homework:
15.5 Applications of Double Integrals
15.6 Triple integrals
15.7 Cylindrical Coordinates; 15.8 Spherical Coordinates Problem 15.5.6:
Find the mass and center of mass of a thin
plate (lamina) occupying the triangular region
with verticies at (0, 0), (1, 1) and (4, 0),
if the density at (x, y ) is ρ(x, y ) = x.
ρ(x, y ) dA, We have mass m =
D and need an iterated integral to evaluate the
double integral. Checking the sketch, we see
that the region is of Type II, with 0 ≤ y ≤ 1,
and y ≤ x ≤ 4 − 3y (where the line from (1, 1)
to (4, 0) has slope − 1 , and we solve
y − 0 = − 3 (x − 4) for x). The iterated integral
4−3y 1 is then x dxdy. Evaluation gives m =
3. y For the center of mass, (x, y ), we have x = 1
m xρ(x, y ) dA,
D 2 and y = 1
m yρ(x, y ) dA. The iterated integrals have
D the same limits as above, but in x the integrand is xρ = x2 ,
and in y the integrand is yρ = xy. The integrals evaluate to 7 and 1,
so dividing by mass, (x, y ) = 1
3 (7, 1) = (2.1, 0.3). Problem 15.6.x Evaluate the iterated integral
z 1 x+z 6xz dydxdz.
0 0 0 Solution. The ﬁrst inside integral has the anti-derivative
6xz dy = 6xyz, so the deﬁnite
6xz dy = [6xyz ]y=0 +z = 6x(x + z ). integral is
0 From this point, the problem is a double iterated integral, with value =1.
Finally, we covered problems 15.7.9 (used for #35) then
15.8.10 and 15.8.18. ...
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