Find the area of the figure bounded by the lines given equations in polar coordinates r=φ and r=2 (Use integrals)
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Have you ever done integrals in polar coordinates before, or is this your first one? – Arthur May 20 '17 at 20:00
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Hello. I need to find the area of this figure using applications of the definite integral, but i don't know how – Макс Мотовилов May 20 '17 at 20:05
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I could tell that from looking at your question. I will now repeat mine: Have you ever done an integral in polar coordinates before, ir is this your first one? – Arthur May 20 '17 at 20:06
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It's my first exp – Макс Мотовилов May 20 '17 at 20:07
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Have you done regular integrals (in $x$-$y$-coordinates)? – Arthur May 20 '17 at 20:07
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@Arthur Yes, of course – Макс Мотовилов May 20 '17 at 20:11
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@Arthur In my case: a=0 and b=2 That's right? – Макс Мотовилов May 20 '17 at 20:23
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I'm writing a full answer; it became much too long for a comment. But yes, that's right. – Arthur May 20 '17 at 20:32
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Here is a short version of general polar coordinate integrals: In regular coordinates, you find area below the graph of a function $y(x)$ between $x = a$ and $x = b$ by approximating the region with very thin rectangles, where the width is represented by $dx$ and the height is $y(x)$. This makes the integral read $\int_a^by(x)dx$, the way you're used to.
In polar coordinates, you can approximate the region by very narrow triangles, with one point in the origin and the two other points along the graph. We let $d\phi$ be the angle at the origin vertex of the triangle. The width of the base of the triangle is then given by (very nearly) $r(\phi)d\phi$.
The height of the triangle is $r(\phi)$so the total area of the triangle is $\frac12(r(\phi))^2d\phi$. This makes the integral of such a region, from $\phi = a$ to $\phi = b$ into $$\int_a^b\frac12(r(\phi))^2d\phi$$where once you've inserted everything in its proper place, you calculate the definite integral the normal way. (The only difference is that the variable is called $\phi$. Everything that has to do with the fact that our coordinate system is polar instead of rectangular was taken care of when setting up the integral, so the calculation works like it always does.)
Since your region is about the area between two graphs, calculate everything inside the outer curve, and then subtract what is inside the inner curve afterwards. The $(r(\phi))^2$ means you can't just subtract the functions and integrate the difference like with regular functions.
Trying to figure out the specifics of your problem, the outer curve is given just by $r_o(\phi) = 2$. We are integrating from $\phi = 0$ to wherever the two curves intersect, which is $\phi = 2$. We then get that the outer area is $$ A_o = \int_0^2\frac12\cdot 2^2\,d\phi = \int_0^22\,d\phi = 4 $$ The inner curve is given by $r_i(\phi) = \phi$, and the inner area is therefore $$ A_i = \int_0^2\frac12\phi^2\,d\phi = \frac43 $$ This makes the total area equal to $$A_t = A_o-A_i = \frac{8}{3}$$
Arthur
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Thanks for the detailed response it turns out it's not hard. And now I understand the difference between polar coordinates – Макс Мотовилов May 20 '17 at 20:51