Answer to Question #185085 in Differential Geometry | Topology for SARALA DEVI

Question

Find the mass of the solid bounded by z = 1 and , z = x² + y² the density function

being d (x, y, z) = | x |

Answer 1

(a). The region for Z 1

Z 1−x

2

1−x

f(x, y) dy dx is bounded by y = 1−x

2 above and by y = 1−x

below. (The two curves intersect at (0, 1) and (1, 0); see me for a sketch.) Solving for x, those two

curves can be written as x = 1 − y (to the left) and x =

√

1 − y (to the right). So the new integral is

Z 1

Z √

1−y

1−y

f(x, y) dx dy.

(b). The region for Z ln 3

Z 3

e

y

g(x, y) dx dy is bounded by x = e

y

to the left and above, by x = 3 to the

right, and by y = 0 below. The curve x = e

y

is of course y = ln x, which intersects y = 0 and (1, 0),

and x = 3 at (3, ln 3). So the new integral is Z 3

1

Z ln x

g(x, y) dy dx.

2. Compute Z Z

D

2x + y dA, where D is the region in the plane bounded by the line y