# Thread: Combining of 2 Double Integrals into 1: how to convert to polar

1. ## Combining of 2 Double Integrals into 1: how to convert to polar

Hello. On this problem shown in the attached image, I have to combine two double integral sums into one double integral.

15. This sum of two double integrals may be written as one double integral.

. . . . .$\color{blue}{ \displaystyle \int_0^3\, \int_0^{\frac{4x}{3}}\, f(x,\, y)\, dy\, dx\, +\, \int_3^5\, \int_0^{\sqrt{\strut 25\, -\, x^2\,}}\, f(x,\, y)\, dy\, dx }$

What is this one double integral?

Answer:

. . . . .$\color{green}{ \displaystyle \int_0^4\, \int_{\frac{3y}{4}}^{\sqrt{\strut 25\, -\, y^2\,}}\, f(x,\, y)\, dx\, dy }$

...or:

. . . . .$\color{green}{ \displaystyle \int_0^{0.927}\, \int_0^5\, f(r,\, \theta)\, r\, dr\, d\theta }$

Looking at the answer, it seems one easy way to do it is to convert to polar coordinates. How would I go about doing this?

2. Originally Posted by TheNerdyGinger
Hello. On this problem shown in the attached image, I have to combine two double integral sums into one double integral.

15. This sum of two double integrals may be written as one double integral.

. . . . .$\color{blue}{ \displaystyle \int_0^3\, \int_0^{\frac{4x}{3}}\, f(x,\, y)\, dy\, dx\, +\, \int_3^5\, \int_0^{\sqrt{\strut 25\, -\, x^2\,}}\, f(x,\, y)\, dy\, dx }$

What is this one double integral?

Answer:

. . . . .$\color{green}{ \displaystyle \int_0^4\, \int_{\frac{3y}{4}}^{\sqrt{\strut 25\, -\, y^2\,}}\, f(x,\, y)\, dx\, dy }$

...or:

. . . . .$\color{green}{ \displaystyle \int_0^{0.927}\, \int_0^5\, f(r,\, \theta)\, r\, dr\, d\theta }$

Looking at the answer, it seems one easy way to do it is to convert to polar coordinates. How would I go about doing this?
Sketch The Areas of the two double integrals first.

3. ## Combining of Double Integrals?

Originally Posted by Subhotosh Khan

Sketch The Areas of the two double integrals first.
I drew it and saw that it was a circle with a radius of 5. (Hence r going from 0-5). Also, y = 4pi/3 intersects this circle at x = 2.73. I tried doing the inverse tangent of y/x and got a theta value of 0.993. This isn't quite the answer that was given. What am I missing? Is there something I need to subtract?

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