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I need to calculate the area by using double integrals.

This is the integral:

\begin{equation*} \iint \limits_{D}xy^2 \,dx\,dy \end{equation*}

This is the domain D: \begin{cases} \ x+y \geq 1 \\[2ex] x^2+y^2 \leq 1 \end{cases} By writing the equations which determine the domain I get this: Circle with the line.

The area in blue must be calculated.

Now i know that x moves along only 0 and 1, I can not figure out the limits for y. I will be thankful if you can help me. The answer provided is: 1/20.

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  • $\begingroup$ The lower limit is given by the line $y=1-x$ and the upper limit by the line $y=\sqrt{1-x^2}$, with $0<x<1$ $\endgroup$ – Jordan Abbott Jun 16 '19 at 20:32
  • $\begingroup$ @CooperCape that gives the right answer, thank you very much, but will you please explain me why is it like that, I am a bit puzzled. $\endgroup$ – tgarmp Jun 16 '19 at 20:37
  • $\begingroup$ By the inequalities given in the question - we have $y\geq 1-x$ and $y^2\leq 1-x^2$ $\endgroup$ – Jordan Abbott Jun 16 '19 at 20:40
  • $\begingroup$ I suppose this works for most of the problems of this kind. I appreciate it. $\endgroup$ – tgarmp Jun 16 '19 at 20:43
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enter image description here

The chord is $\;y=1-x\;$ , the circle is $\;x^2+y^2=1\;$ , and you want the part on the first quadrant between the chord and the circle.

$$\int_0^1\int_{1-x}^{\sqrt{1-x^2}}xy^2\,dy\,dx=\frac13\int_0^1x\left((1-x^2)^{3/2}-(1-x)^3\right)dx=$$

$$=-\left.\frac16\frac25(1-x^2)^{5/2}\right|_0^1-\frac13\int_0^1x(1-3x+3x^2-x^3)\,dx=$$

$$=-\frac1{15}(0-1)-\frac13\left(\frac12-1+\frac34-\frac15\right)=\frac1{15}-\frac16+\frac13-\frac14+\frac1{15}=$$

$$=\frac{8-10+20-15}{60}=\frac1{20}$$

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  • $\begingroup$ May I ask you how you made the calculations so fast? $\endgroup$ – tgarmp Jun 16 '19 at 20:45
  • $\begingroup$ @tgarmp The last ones? Simple arithmetic For the integrals it is almost immediate: $$\int x(1-x^2)^{3/2}\,dx=-\frac12\int d(1-x^2)\,(1-x^2)^{3/2}$$ and etc. $\endgroup$ – DonAntonio Jun 16 '19 at 20:48
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We require $$D = \{(x,y) \in \mathbb R : x^2 + y^2 \le 1 \le x+y\}.$$ Noting that this implies $x \ge 0$ and $y \ge 0$, we can solve the associated inequalities for $x$ as a function of $y$ on the interval $y \in [0,1]$: $$\sqrt{1-y^2} \le x \le 1 - y.$$ Consequently, $$\iint_D xy^2 \, dx \, dy = \int_{y=0}^1 \int_{x=\sqrt{1-y^2}}^{1-y} x y^2 \, dx \, dy.$$

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I would convert to polar. Then the region becomes:

$r = [\frac{\sqrt{2}}{2}\sec (\theta - \frac {\pi}{4}),1]$

or $r = [\frac {1}{\cos\theta + \sin\theta},1]$

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  • $\begingroup$ That region you depict there, which I cannot understand, has $\;\theta\;$ all around the circle. Yet the region exists only on the first quadrant... $\endgroup$ – DonAntonio Jun 16 '19 at 21:02
  • $\begingroup$ Ah, sorry, I gave you the complimentary region. I will edit. $\endgroup$ – Doug M Jun 16 '19 at 21:04

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