MIDDLE EAST TECHNICAL UNIVERSITY
DEPARTMENT OF MECHANICAL ENGINEERING
ME 210 APPLIED MATHEMATICS FOR MECHANICAL ENGINEERS
SPRING 2014
HOMEWORK 6 COMMENTS AND COMMON MISTAKES
Question 3.
Given I =
 (3x  z)dV
where V is the volume enclosed between the paraboloid y  x2  z2 and
V
the plane y=5
a. State the integral I in the integration order, dxdydz expressing the integral limits clearly. Do not
evaluate the integral.
b. State the integral I in the integration order dzdxdy expressing the integral limits clearly. Do not
evaluate the integral.
c. State the integral I in the integration order dydxdz. Evaluate this integral.
Common Mistakes:
Parts a&b. Half of the students did not understand how to express a volume integral, how to reduce it
to a double integral and how to find the limits of the three nested integrals.
Your limits for the first integral (the innermost integral) must be equations of surfaces of the volume.
When projecting your volume to a principal plane (to perform the next two integrations, i.e. the double
integral), always plot the projected area on the projection plane so that you can understand the region
R of double integration (e.g. For dxdydz order, your projection plane will be the y-z plane and the
projection of volume V on yz-plane will be a region R on this plane). After plotting region R,
determine the limits of the double integral (the outer two integrals) using the boundary curves of the
projected area R. Based on your integration order of the double integral (after you have integrated the
innermost integral), the limits of the middle (inner) integral may include the coordinate variable
according to which the final (outer) integral is integrated. For example, for a volume integral of
dxdydz order, your first (innermost) integral limits may contain x and y but not z (integration with
respect to z), the inner (middle) integral limits may contain x (but not y or z) and the last (outermost)
integral limits will be constants.
c. All of issues mentioned for parts a & b on writing the limits of the integrals are valid for this part as
well. Additionally, some of you did not work on the analytical solution of the integral. After showing
some initial steps, some of you have used a calculator or computer software to integrate, but this was
an integral that had to be evaluated by hand. Points were deducted because of this. Please keep in mind
that in the examination, all integrations will be performed by hand and all steps should be presented
clearly.
Problem 6.
a. Verify Stokes’ theorem by calculating the curl and circulation integrals
  F  ndS   F  dr
S
C
1
where F  yzk and S is the surface z  1  2x2  y2 for x  0 , y  0 and z  0 .
2
1
b. Verify Stokes’ theorem using the same vector function F  yzk , the same circulation path C but
2
another surface S.
ME210/2014Spring/HW6/Comments and Common Mistakes
Page 1/2
MIDDLE EAST TECHNICAL UNIVERSITY
DEPARTMENT OF MECHANICAL ENGINEERING
ME 210 APPLIED MATHEMATICS FOR MECHANICAL ENGINEERS
SPRING 2014
c. If the unit vector n were taken in the opposite direction to that in your solution of part (a), how
should the surface integral expression and the circulation path in line integral change so that the
Stokes’ theorem holds? In that case, what would be the result of each (line and surface) integral?
Explain/show clearly.
Common Mistakes:
a. Most of the students did not show the direction of the unit normal and the direction of circulation.
You had to show that the unit normal and the orientation of the circulation path are consistent.
Some of the students did not divide the normal vector by its magnitude, while evaluating the unit
normal vector.
The surface integral had to be calculated as follows:
 
S

1
  F  ndS
2
 
S
2
 g   g 
  F  n 1       dxdy
 x   y 

2
However, most of the students just wrote dS  dxdy , which is not correct.
You had to show that
  F  ndS   F  dr
S
in order to verify Stokes’ theorem. However, most of
C
the students did not calculate the right hand side of the equality.
b. In this part, you were asked to verify Stokes’ theorem using the same vector function, same
circulation path but another surface S. Some students simply wrote that the result is the same as part
(a) because Stokes’ theorem holds, which is a correct comment but not the verification of Stokes’
theorem. You should have defined another surface, calculated the left hand side of the equality using
the new surface and shown that the result was equal to the value found in part (a).
You should have also shown that the unit normal vector of the new surface was consistent with the
orientation of the circulation path.
c. Some of the students just wrote that the result would be the negative of the value found in part (a)
and (b). This comment is not enough since it was asked to explain/show clearly how the surface
integral expression and the circulation path in line integral should change so that Stokes’ theorem
holds. In that case, what would be the result of each (line and surface) integral?
ME210/2014Spring/HW6/Comments and Common Mistakes
Page 2/2