It is known that in the cuboid ABCD A1B1C1D1, AB 3, AD 4, BB1 5, find the opposite plane line DA1 an

In fact, the triple integral is an extension of the first and double integrals.

Triple integrals and their calculations.

First, the concept of triple integrals.

The definition of triple integral is obtained by generalizing the integral region in the definition of double integral to the spatial region and the integral function to the ternary function.

where dv is called a volume element, and other terms are the same as double integrals.

If the limit exists, the function is said to be integrable.

If the function is continuous over a closed region, it must be integrable.

This is known by definition.

Triple integrals have exactly the same properties as double integrals.

The physical background of the triple integral.

The mass of a space object with f ( x, y, z ) as the bulk density.

Let's discuss the calculation method with the help of the physical background of triple integral.

Second, the calculation method in the Cartesian coordinate system.

If we divide the spatial region with the three planes x = constant, y = constant, z = constant, then each regular small region is a cuboid.

Its volume is , so the area element in the Cartesian coordinate system is .

Triple integrals can be written as:

Similar to double integrals, triple integrals can be converted into cubic integrals for calculation.

Specifically, it can be divided into single before heavy and heavy first and heavy after single.

First single and then heavy — also known as first one and then two, the strip method (first z, then y, then x).

Note: Exactly the same method can be used to convert triple integrals into cubic integrals in other sequences.

Steps to integrate cubic

projection, get the flat area.

Crossing the legal limit, the entry point - the lower limit, the exit point - the upper limit.

For double integrals, we have already introduced the method of turning into progressive integrals.

Example 1 will be converted into cubic integrals.

where is a cuboid, and each boundary plane is parallel to the coordinate plane.

The solution projects d onto the xoy plane, which is a rectangle.

Fix any point (x,y) in d as a straight line parallel to the z-axis.

The boundary surface intersects at two points with vertical coordinates of l and m (l < m) o x

y z m l

a b c d

d (x,y)

Example 2 Calculation.

where is the area d x enclosed by the three coordinate planes and the plane x + y + z =1

y z o solution.

Draw the solution of the region d In addition to the single-second-repeat method introduced above, the triple integration can also be converted into cubic integration using the first-repeated, single-repeated, or slicing method.

Repeat first and then single, that is, first find the double integral about one of the two variables and then find the definite integral about the other variable.

If f(x,y,z) is continuous on .

Between two parallel planes z = c1 , z = c2 (c1 < c2).

Use any plane parallel to the two planes to intercept the area.

Then the first heavy and then the single.

It is easy to see that if the integrand function is not related to x and y, or the double integral is easy to calculate, it is more convenient to use the cross-section method, that is, the area of the cross-section, such as a circle, ellipse, triangle, square, etc., the area is easier to calculate.

Especially when f ( x , y , z ) is not related to x , y.

Hope it helps.

2 The straight line AC1 is on the plane AA1C1C, and the straight line BB1 is on the plane AA1C1C, and the distance between the straight line BB1 and AC1 is the distance from the straight line BB1 to the plane AA1C1C AA1 AC, BB1 AA1, AC, AA1 on the plane AA1C1C, BB1 is not on the plane AA1C1C BB1 AC The distance between the straight sock slip line BB1 and the plane AA1C1C.

AE plane BCE: point E is the midpoint of A1B1, AA1=A1E=A, AE BE is obtained, and BC plane bright slag ABB1A1 is obtained, BC AE, BC, BE are the two intersecting edges of the bond skin on the plane BCE, and AE plane BCE is obtained. Banquet difference.

Connecting C1F, C1D, C1F AC, C1D AE (obtained from C1F plane ABCD, C1D plane abb1A1), according to the two intersecting lines in one plane parallel to the two intersecting lines in the other plane, then the two planes are parallel. Plane C1DF Plane ace, DF belongs to the straight line on plane C1DF, and DF plane ACE is obtained.

With DA as the X-axis, DC as the Y-axis, and DD1 as the Z-axis, a Cartesian coordinate system is established. Grinding judgment.

Mark the coordinates of these slippery points. Let's hole.

It's easy to figure it out ...

Take the midpoint M on B1C1 and connect EM and CM, then AEMC is a plane, and it is easy to verify that DF is parallel to CM, and CM is in the Pingque Liang Stool is in the ACE of the travel surface, then the DF parallel scum surface ACE

Connect CE, because dd1=a, d1e=a, so de=2 (1 2)a, the same way can be balanced ec=2 (1 2)a, and because cd=2a, so ascend next to de ce. Because the BC face C1CD, so the BC DE. So de flat noisy block oak noodle bce

Because cd is perpendicular to the face add1a1

So CD is perpendicular to AD1

Because ad=aa1 and add1a1 is rectangular.

So AD1 is perpendicular to A1D

And because CD is perpendicular to AD1

So AD1 is dull and the segment is straight plane A1DC

It is the volume of the cuboid that is half minus the volume of e-aa1b1b, which is equal to 2 3

Solution: If CD1 is connected, then Ba1 Cd1, B1CD1 is the angle formed by two straight lines from different planes, in B1CD1, by AB=BC=1, AA1=2, B1C= 5, CD1= 5, AD1= 2

cos∠b1cd1=5+5-22×5=45

Therefore, choose B