- #1
iScience
- 466
- 5
consider the following image
(the red is the surface area element and the green is the differential element that I'm integrating over)
when we derived this in class, we treated the area formed by vectors a and b, as the area of a parallelogram. the thing is, a and b should be at right angles implying that rest of the angles should be right angles as well. if that's the case, then why can't i just treat the area formed from a and b, as a rectangle?
well, that's what i attempted to try but failed. i get all the appropriate terms as they do in the surface area formula, except i get an extra term as shown in the link below, implying, perhaps.. that that extra term must go to zero. but i don't know why it should.
if I'm incorrect in treating the shape formed from a and b as a rectangle, why is this incorrect?
calculations:
http://i.imgur.com/6MFO4xI.jpg
(image was too large to put here)
(the red is the surface area element and the green is the differential element that I'm integrating over)
when we derived this in class, we treated the area formed by vectors a and b, as the area of a parallelogram. the thing is, a and b should be at right angles implying that rest of the angles should be right angles as well. if that's the case, then why can't i just treat the area formed from a and b, as a rectangle?
well, that's what i attempted to try but failed. i get all the appropriate terms as they do in the surface area formula, except i get an extra term as shown in the link below, implying, perhaps.. that that extra term must go to zero. but i don't know why it should.
if I'm incorrect in treating the shape formed from a and b as a rectangle, why is this incorrect?
calculations:
http://i.imgur.com/6MFO4xI.jpg
(image was too large to put here)