Unit 6: Integral Calculus - Subjective Questions

Definition: A double integral of a function over a region is written as It represents the limiting sum of the values of over small elements of area.

Geometrical interpretation:

• If , the double integral gives the volume under the surface and above the region in the -plane.
• If , it gives the area of the region .

The given integral is

First integrate with respect to :

Now integrate with respect to :

Therefore, the value of the double integral is

An iterated integral is a double integral evaluated successively with respect to one variable and then the other. In the inner integral is evaluated first with respect to , treating as constant.

For the given integral,

Integrating with respect to gives

Thus,

Therefore,

Hence,

The given integral represents the region

The region is bounded by:

• ,
• ,
• .

When the order is changed, varies from to . For a fixed value of , varies from to . Therefore, the reversed order is

Changing the order is useful when:

• The inner integral becomes easier to evaluate.
• The original order contains difficult limits or an unevaluated function.
• The region has a simpler description in the reversed order.

The region is given by

Since , we have . The limits in the reversed order are

Therefore,

Evaluating the inner integral,

Thus,

The original region is

The curve is , or equivalently in the first quadrant. The total range of is .

For , the condition gives

However, the boundary is reached when , so the region does not require a split in this particular case. Hence the changed order is

In general, a region must be split when the upper or lower boundary for the inner variable changes at an intermediate value of the outer variable. This occurs when one vertical or horizontal line intersects different boundary curves in different parts of the region.

Suppose the transformation is

The Jacobian of the transformation is

It is given by

The change-of-variables formula is

The absolute value of the Jacobian accounts for the change in area. Thus,

The transformation should be one-to-one in the region, except possibly on boundary points.

Use the transformation

The boundary curves become

Solving for and ,

The Jacobian is

Thus,

Since ,

Therefore,

Hence,

The polar transformation is

Also,
and the area element becomes

For the circle ,

Therefore,

Thus,

Hence,

The area of a region in the -plane is obtained by integrating the elementary area over the region:

The curves intersect when

so

For , the upper curve is and the lower curve is . Therefore,

Evaluating,

Hence,

The circle is

The line divides the circle into a small circular segment on the right and the remaining region on the left. For the region to the right of the line,

with

Therefore, the required area is

This becomes

Using

we obtain

Thus, the area of the smaller segment is

If a surface lies above a region and , then the volume under the surface is

The triangular region is described by

Therefore,

First,

Thus,

Letting gives

Hence, the volume is cubic units.

The main differences are:

• Double integral:
  
  It is evaluated over a two-dimensional region in the -plane. The differential element may be written as , , or . It is used to calculate areas, volumes under surfaces, mass of laminae, and related quantities.

• Triple integral:
  
  It is evaluated over a three-dimensional solid . The differential element is
  
  or another order of three differentials. It is used to calculate volumes, mass of solids, moments, and centers of mass.

• For area:
  

• For volume:
  

Thus, a double integral accumulates quantities over area, while a triple integral accumulates quantities throughout a volume.

A triple integral of a function over a solid region is written as

It represents the accumulation of the function throughout a three-dimensional region.

For the given rectangular solid,

Integrating with respect to ,

Integrating with respect to ,

Finally,

For a rectangular parallelepiped, the triple integral can be evaluated as an iterated integral:

For ,

Since the variables are separable,

Therefore,

Hence,

The tetrahedron lies in the first octant. Solving the plane for gives

The projection on the -plane is the triangle

Therefore,

The inner integration gives

After evaluating the remaining integrals,

This agrees with the standard formula for a tetrahedron whose mutually perpendicular intercepts are , , and .

In cylindrical coordinates,

The cylinder is described by

Hence,

Evaluate successively:

Therefore,

Spherical coordinates are defined by



where:

• is the distance from the origin,
• is the angle measured from the positive -axis,
• is the azimuthal angle in the -plane.

The volume element is obtained from the Jacobian of the transformation:

For the sphere , the limits are

Thus, the volume can be expressed as

This evaluates to

For a sphere of radius , spherical coordinates are appropriate because the equation becomes simply

The complete angular ranges are

The volume element is

Therefore,

Separating the integrations,

Thus,

The surfaces intersect when

which is the circle in polar coordinates.

The plane is above the paraboloid, so the volume is

Using polar coordinates,

with

Hence,

Evaluate the radial integral:

Therefore,

Let a solid be bounded below by and above by , where over the projection in the -plane.

A vertical column above an area element has height

Therefore, the volume is

If the lower surface is the -plane, then , and

Alternatively, the volume may be written as a triple integral:

The projection method is useful because it reduces a three-dimensional volume problem to a double integral over a two-dimensional region. The principal steps are:

• Determine the projection .
• Identify the upper and lower surfaces.
• Set up the height function.
• Evaluate the resulting double integral.