Double integral

[canvas]

[math]\text{Double integral in polar coordinate system}\\\\ \iint_D(x^2+y^2)dxdy,\,\,\,D:\,x^2+y^2\leq1\leq x+y\\\\\text{I need to find range of r so:}\\\\(1)\,\,x^2+y^2\leq1 \Rightarrow\,\,r^2\leq1\,\,\Rightarrow\,\,r\in[0,1]\\\\(2)\,\,x+y\geq1\,\,\Rightarrow r\cos\varphi+r\sin\varphi\geq1\,\,\Rightarrow r(\sin\varphi+\cos\varphi)\geq1\\\\\text{I'm confused how to solve it for r?}[/math]

 

[canvas]

[math]\text{Alternate form of inequality is:}\\\sqrt{2}r\sin\left(x+\frac{\pi}{4}\right)\geq1[/math]

 

[blamocur]

First integrate with respect to [imath]r[/imath] over the segment which depends on [imath]\phi[/imath], then integrate with respect to [imath]\phi[/imath].

 

[Deleted member 4993]

[math]\text{Double integral in polar coordinate system}\\\\ \iint_D(x^2+y^2)dxdy,\,\,\,D:\,x^2+y^2\leq1\leq x+y\\\\\text{I need to find range of r so:}\\\\(1)\,\,x^2+y^2\leq1 \Rightarrow\,\,r^2\leq1\,\,\Rightarrow\,\,r\in[0,1]\\\\(2)\,\,x+y\geq1\,\,\Rightarrow r\cos\varphi+r\sin\varphi\geq1\,\,\Rightarrow r(\sin\varphi+\cos\varphi)\geq1\\\\\text{I'm confused how to solve it for r?}[/math]

I would

first sketch the circle x² + y² = 1 and​

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the line x + y = 1​

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locate the points of intersection and hatch the the zone of integration​

Now integrate

 

[canvas]

First integrate with respect to [imath]r[/imath] over the segment which depends on [imath]\phi[/imath], then integrate with respect to [imath]\phi[/imath].

I know how to integrate, but I don't know the limits of integration of r ;/

 

[Deleted member 4993]

I know how to integrate, but I don't know the limits of integration of r ;/

Have you sketched the area of integration? That will help you to establish the limits.

 

[Dr.Peterson]

I know how to integrate, but I don't know the limits of integration of r ;/

Have you written the polar equation of the line? Yes, you did, right at the start. Solve that for r as a function of [imath]\phi[/imath]. Since r has to be greater than that, that's your lower limit.

Or are you, contrary to #3, planning to integrate with respect to [imath]\phi[/imath] first? It will help a lot if you show the integral you've set up, so we can be sure what limits you are looking for.

 

[blamocur]

I know how to integrate, but I don't know the limits of integration of r ;/

Limits -- actually only one of the limits -- of integration for [imath]r[/imath] will depend on [imath]\phi[/imath]. Can you figure out that dependency?

 

[nasi112]

[math]\text{Double integral in polar coordinate system}\\\\ \iint_D(x^2+y^2)dxdy,\,\,\,D:\,x^2+y^2\leq1\leq x+y\\\\\text{I need to find range of r so:}\\\\(1)\,\,x^2+y^2\leq1 \Rightarrow\,\,r^2\leq1\,\,\Rightarrow\,\,r\in[0,1]\\\\(2)\,\,x+y\geq1\,\,\Rightarrow r\cos\varphi+r\sin\varphi\geq1\,\,\Rightarrow r(\sin\varphi+\cos\varphi)\geq1\\\\\text{I'm confused how to solve it for r?}[/math]

You have already solved for \(\displaystyle r\).

\(\displaystyle \frac{1}{\sin \theta + \cos \theta} \leq r \leq 1\)


\(\displaystyle \int_0^1 \int_{1-y}^{\sqrt{1-y^2}} (x^2 + y^2) \ dx \ dy \ = \int_0^{\frac{\pi}{2}} \int_{\frac{1}{\sin \theta + \cos \theta}}^1 r^3 \ dr \ d\theta\)

 

[canvas]

Thank you guys, I got it now, was so easy but I'm blind sometimes...
What about next integral, integration area is so weird...
[math]\iint_D(x+y)dxdy,\,\,D:\,\,x\geq0,\,\,y\geq0,\,\,\sqrt{x}+\sqrt{y}\leq1[/math]

 

[nasi112]

The polar integral is easy to compute.

 

[canvas]

The polar integral is easy to compute.

But how to find range of [math]\varphi \,\,\text{and} \,\,r[/math]??

 

[nasi112]

\(\displaystyle \frac{1}{\sin \theta + \cos \theta} \leq r \leq 1\)

And

\(\displaystyle 0 \leq \varphi \leq \frac{\pi}{2}\)