Recent content by CallMeShady

I can calculate the area exactly without rotation; however, I could not think of a way to find the centroid of the triangle in the given orientation in the question. Hence, by making a simple rotation, it just made my life easier.

And as haruspex mentioned, the formula I was using did not rotate the co-ordinate counterclockwise. A quick Google search led me to the correct formula, and now I can satisfyingly say that I found the correct answer: I = -7.027811891 It's off by 0.01 from one of the answers, but that should be...

The centroid co-ordinates are (1.110899311, 0.555130063). The first co-ordinate is for the base. Basically what I did was position the right triangle's base on the x-axis, and then find the centroid co-ordinates of it. Then I rotated the triangle by 26.5 degrees counter clockwise in order to...

http://i62.tinypic.com/zxtvsx.jpg Might be hard to read, but maybe it may be visible enough? Please click on the image, and then select the option (view raw image) to zoom in.

Ahhhhh this is going to make me have nightmares tonight! So I found a calculation error in determining the height of the right triangle, so I re-did all the calculations with the correct values. I got an answer of -2.29. It's not one of the answers. Then I rounded my values of base, height, and...

Interesting stuff. So I broke up the integral and recognized this: I = a∫∫dA + b∫∫xdA + c∫∫ydA. The first terms is simple "a" times the area, the second term is the centroid of x times the area times b, and the last terms is the centroid of y times the area times c. My centroid points appear...

Well, for a right triangle, it's simple as the formulas are already dervided online. Or, we could simply integrate using (assuming density is constant): x_centroid = 1/A * ∫∫x*dA y_centroid = 1/A * ∫∫y*dA Suppose I do find the centroid co-ordinates, how does that result into determining a...

No, I don't recognize this from any of my courses. I was able to follow through the logic that you presented. However, shouldn't it be ##\vec{u} = (b,c)##? Also, when you break the integral into two in the end, don't I still have to use anti-derivatives for the second piece in x and y (as...

Well, I am still a bit confused as to where to go from here. After reading the responses and doing some research online, I stumbled upon something called the "Jacobian." Am I drifting off on a tangent to infinity?

Yes, the values are correct (as shown in my question). Althought I must point out that two previous questions have had incorrect sets of answers in this particular problem set. My professor acknowlegded them as well.

Hm... it represents a plane. Are you suggesting that I find the volume of the two pieces (triangular base + another triangular piece on top with the slanted plane)?

Unfortunately, I don't quite understand the idea that you are suggesting. I am unsure on how switching two co-ordinates will achieve anything in regards to this problem.

Homework Statement Make a good sketch of the plane region D defined by the following simultaneous inequalities: D: y >/= -2x, 2y >/= x, 2y </= 4-x. Use deep conceptual understanding the insight (and no antiderviative calculations!) to reduce the iterated integral below to a simple algebraic...

Homework Statement The Diesel cycle is an idealized representation of the process that occurs in a Diesel combustion engine, as shown in the graph below. Starting at point a, the Diesel cycle consists of an adiabatic compression, an isobaric expansion, and adiabatic expansion and an isochoric...

I am not sure what you mean with that equation above... what does the hat above the variables mean? After substituting the numbers, I got r = √35 and plugging that into the original derived equation, I got an answer of -7.997 \times10-6 (which I am assuming is an answer in Newtons).