![]() ![]() Minus two cosine theta plus cosine squared theta, d theta. So this is going to be equal to one half times the definite integralįrom zero to two pi. Point you feel inspired, try to evaluate this. And now we just have toĮvaluate this integral. Now what's r of theta squared? Maybe I'll color code this a little bit. Is equal to zero radians to theta is equal to two pi radians. Pi, cosine of two pi is one, one minus one is zero again. And then we go all the way around to theta is equal to two pi radians. Zero radians is 1 minus 1 we're right over there. Is equal to zero radians, and we're essentially going all the way- When theta is equal to Now what's our alpha and what's our beta? Well, we're going from theta Is going to be equal to one half the definite integral. Just have to apply this to this function right over here. To our ending theta, from alpha to beta, of r Half the definite integral from our starting theta So we've already seen, we'veĪlready given ourselves the intuition for the formula, that the area enclosed by a polar graph is going to be equal to one And I encourage you to pause the video and try it on your own. ![]() ![]() To see if we can figure out the area enclosed by this curve. To 1 minus cosine of theta, of course we're dealing So this darker curve in blue is the graph of r is equal I hope this helps some! It's definitely an easy mistake to make, so I think the best advice I can give is try to reason through it mathematically, not graphically, as the graph will sometimes trick you. Because of this, you must use 0 to 2π as your bounds of integration, otherwise you will cut off this tiny tiny piece on the positive x-axis, and your area will be wrong. For this graph, if you were to zoom in very closely (try plotting this graph on if you would like) you would see that the graph just barely grazes the positive x-axis. This is also much better than eyeballing the bounds of integration, which is very tempting for curves like these, but often can result in wrong answers due to very small regions where the function crosses a certain axes, or achieves a certain value that is very difficult to see just by eyeballing it. Sometimes I like to do this without looking at the graph first because then I don't have any mental picture of what I think the bounds should be that might mess up my calculations. This will give you candidates for the bounds of integration. I've found that the best way to be sure that your bounds of integration are correct is to set the expression r(□) equal to zero. This is definitely a mistake I've made on several problems. ![]()
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