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Email this article Login required. Email the author Login required. Hide Show all. Open Journal Systems. User Username Password Remember me. Language Select Language English Indonesian. Abstract The aim of this research is to describe crithical thinking ability of the students with low problem solving ability in solving problem in Riemann Integral based on the criteria of critical thinking in problem based learning.

The data collection used is test and interview. Based on the result of this research can be known that profile of critical thinking ability with the low level of problem solving ability was can do basic clarification, can not do bases for a decision, can not do inference, can not do advance clarification, and can not do supposition and integration.

Keywords mathematical critical thinking ability, mathematical problem solving, problem based learning, integral calculus. Full Text: References Aizikovitsh-Udi, E. Diakses Facione, P. International Journal of Education, 4, 1 , Moleong, J.

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Problem Set Because the slopes of both roofs are the same, the area of Tm will be the same for both roofs. Therefore, the area of the roofs will be 2a the same.

Thus the surface b. The surface area of a paraboloid and a hyperbolic surface and makes an acute angle with the z-axis. Then the normal So, the areas depend on the regions. Then obtain that its equation is See note with next Note that this is 0 not valid if we are concerned with values of or moments or mass. See note with previous The moment of inertia with respect to the y-axis is the integral over the solid of the function which gives the square of the distance of each point in the solid from the y-axis.

It will be helpful to first label the corner points at the top of the region. The resulting projection is shown in the figure above and to the right. The possible values of x depends on where we are in the yz-plane. Therefore, we split up the solid into two parts. The volume of the solid will be the sum of these two smaller volumes.

Figure 1: When the center of mass is in this position, it will go lower when a little more soda leaks out since mass above the center of mass is being removed. Figure 2: When the center of mass is in this position, it was lower moments before since mass that was below the center of mass was removed, causing the center of mass to rise.

Therefore, the center of mass is lowest when it is at the height of the soda, as in Figure 3. The same argument would hold for a soda bottle.

The result obtained from a CAS is: The region is a right circular cylinder about the z- The region is a hollow right circular cylinder 6. The region is one-eighth of a sphere in the first about the z-axis with inner radius 1, outer radius octant of radius a, centered at the origin. See comment at beginning of write-up of Assume that the hemisphere lies above the xy- Problem 26 of the previous section. Assume that the hemisphere lies above the xy- Consider the following diagram: