Question

evaluate the triple integral. e (x − y) dv, where e is enclosed by the surfaces z = x2 − 1, z = 1 − x2, y = 0, and y = 8

Answer 1

The triple integral

∭

�

(

�

−

�

)

 

�

�

∭

E

​

(x−y)dv, where

�

E is enclosed by the surfaces

�

=

�

2

−

1

z=x

2

−1,

�

=

1

−

�

2

z=1−x

2

,

�

=

0

y=0, and

�

=

8

y=8, evaluates to

−

256

15

−

15

256

​

.

Step-by-step explanation:

To evaluate the triple integral, we need to set up the integral over the region

�

E defined by the given surfaces. The bounds for

�

x,

�

y, and

�

z are determined by the intersection points of these surfaces. Since

�

y ranges from 0 to 8,

�

x is bounded by the curves

�

=

�

2

−

1

z=x

2

−1 and

�

=

1

−

�

2

z=1−x

2

. Solving these equations provides the bounds for

�

x.

The triple integral becomes

∫

0

8

∫

−

1

+

�

1

+

�

∫

�

2

−

1

1

−

�

2

(

�

−

�

)

 

�

�

 

�

�

 

�

�

∫

0

8

​

∫

−

1+z

​

1+z

​

​

∫

x

2

−1

1−x

2

​

(x−y)dydxdz. Solving this iteratively, we find that the final answer is

−

256

15

−

15

256

​

after performing the necessary calculations.

In summary, the evaluation involves setting up and solving a triple integral over the specified region

�

E. Careful consideration of the bounds based on the given surfaces leads to the result

−

256

15

−

15

256

​

.

Answer 2

To evaluate the triple integral e^(x-y) dv, you need to set up the integral using the bounds provided by the intersecting paraboloids and the given y range. The proper limits for x, y, and z should be determined, and the integration should be performed in the correct order, applying relevant techniques.

Step-by-step explanation:

The question involves evaluating a triple integral of the function e(x-y) over a volume in three-dimensional space bounded by specified surfaces. To approach this, you need to set up the integral using the given bounds. The surfaces are paraboloids defined by z = x2 − 1 and z = 1 − x2, with y ranging from 0 to 8. These bounds imply that the volume of integration is symmetric about the z-axis and lies between these two paraboloids.

Next, determine the intervals for x, y, and z for the triple integral. Notice that the paraboloids intersect when x2 - 1 = 1 - x2, so solving for x gives x = 0. Thus the region is limited to between x = -1 and x = 1 since it is symmetric about the z-axis. Then, we must consider the bounds for z, which will be from the lower surface z = x2 - 1 to the upper surface z = 1 - x2, and y varies from 0 to 8 as given.

You would perform the integration with respect to z first, followed by x, and finally y, applying the appropriate limits for each variable. Keep in mind proper application of integration techniques, such as performing substitutions if necessary, to solve the integral.