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junior college 1 | H2 Maths
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Not sure how to do this.. tks!
Date Posted: 2 years ago
Views: 369
The formula for area under a curve is
(Integrate sign) y dx
and then we need to replace the y and dx by their equivalents in theta and dtheta
Will look at this later if I can remember
(Integrate sign) y dx
and then we need to replace the y and dx by their equivalents in theta and dtheta
Will look at this later if I can remember
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Even though the actual ellipse appears on both the positive and negative portions of each axis (appearing on all four quadrants), we are doing the integral parametrically.
In O Level A Maths, you may have recalled that we have to be careful when calculating areas which appear above and below the x axis.
Here, the curve is defined in terms of rotation. If you actually play around with the limits, you will notice that the shape of the curve is built from the positive x-axis, in an anticlockwise manner, all the way until the shape is completed (when theta = pi). A full shape is enclosed, so we know that the integral we calculated is correct.
If we were to change the upper limit to say theta = 2pi, then we would be enclosing the ellipse twice. Our integral value will eventually be twice of what we are calculating here, so it’s not good.
My point here is that we do not have to break the integral like my first method; in fact, because theta = 0 to theta = 2pi encloses the shape exactly, we can just take the integral and obtain that value as our area.
There is actually another method by really manually converting the parametric equation into a Cartesian form (which is of course going to be that of an ellipse), but you have to be careful with the negative portions of the graph in this case.
In O Level A Maths, you may have recalled that we have to be careful when calculating areas which appear above and below the x axis.
Here, the curve is defined in terms of rotation. If you actually play around with the limits, you will notice that the shape of the curve is built from the positive x-axis, in an anticlockwise manner, all the way until the shape is completed (when theta = pi). A full shape is enclosed, so we know that the integral we calculated is correct.
If we were to change the upper limit to say theta = 2pi, then we would be enclosing the ellipse twice. Our integral value will eventually be twice of what we are calculating here, so it’s not good.
My point here is that we do not have to break the integral like my first method; in fact, because theta = 0 to theta = 2pi encloses the shape exactly, we can just take the integral and obtain that value as our area.
There is actually another method by really manually converting the parametric equation into a Cartesian form (which is of course going to be that of an ellipse), but you have to be careful with the negative portions of the graph in this case.
Date Posted:
2 years ago
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