Aktug Y.

asked • 08/25/20

calculate the integral.

∫02∫0√(2x-x^2) xdydx

Kevin S.

tutor
This is similar to the question from yesterday, except simpler. Did you understand that solution? What have you tried on this one?
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08/25/20

Aktug Y.

Actually, i trying to integrate the inside first, but i don't know how to integrate x with respect dy. I feel like an idiot. Sorry.
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08/25/20

Kevin S.

tutor
No, I didn't mean it that way. But you may get more help if you share what you've tried and where you're stuck. And you didn't comment on the previous solution, so maybe you had a different kind of question. --- If you're integrating wrt y, remember everything else is a constant. y sees x as just a number, so when you integrate it, you get xy, just like if you integrate 2, you get 2y.
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08/25/20

Aktug Y.

now i can't integrate x.√((x-2).x)) wrt x. i think something going wrong. should i try x-u transformation or what should i do?
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08/25/20

Kevin S.

tutor
Ok good. You made it that far. I see what this is tricky now. You'll need to know a specific trick. I'll show you below. Give me a minute to write it.
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08/25/20

Kevin S.

tutor
I fixed my solution below. When you sub sine in for something, the domain should always be [-π/2, π/2] to keep thing invertible.
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08/25/20

Aktug Y.

Kevin, thank you for your kindess and all of your help.
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08/26/20

1 Expert Answer

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Ryan B. answered • 08/25/20

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So there is the way to solve it done by Kevin, which 100% does work! However we can solve it via transforming the formula from the Cartesian plane to the polar plane. The outer integral would be based on the angles (d∅) of the graph, which go from 0 to pi/2. The region of integration is the top half of the disk centered at (0, 1) with radius 1. To see this, observe that the upper bound on y is √ 2x − x 2. The inner integral would be the radius or distance dr. The bottom would remain 0. The top would be transformed using x=rcos∅ and y=rsin∅. You find that r=2cos∅. The interior function would be r2cos∅.

The interior integral would be from 0 to 2cos∅ of r2cos∅ dr. So 1/3r3cos∅. (2cos∅)^3 = 8cos3∅. Which would mean that the new expression would be 8/3cos4∅.


Integrating that gives you the indefinite of 8/3( (sin(4∅)+8sin(2∅)+12∅)/32). Finding the definite you plug in and have (after all the work) 8/3 * 3pi/16 which yields pi/2 as the final answer.

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Aktug Y.

Thank you so much. What if we do x-u transformation? Can we find the answer with this method?
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08/25/20

Kevin S.

tutor
What happened to the sqrt in the integral bounds? I don't think that will be the answer.
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08/25/20

Ryan B.

...it might have just disappeared haha, thanks for catching that
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08/25/20

Ryan B.

I'll edit once I am back to my computer to fix it to the correct answer
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08/25/20

Kevin S.

tutor
It's a doozy. I hit a snag picking the wrong bounds for the trig sub. I'd like to see a cleaner way if you have it.
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08/25/20

Ryan B.

HAHA, Polar Coordinates!
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08/25/20

Kevin S.

tutor
Nevermind, I got it below.
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08/25/20

Aktug Y.

I can't thank you guys enough, especially Kevin. Thank you very much for your help.
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08/26/20

Kevin S.

tutor
Oh, no, Ryan's is way better. I can't believe I didn't see that. Coordinate transformations are much more important for domain shape than the integrand. (Although I think there's value in seeing that you can push through an integral even if you don't know the "right" way.) Also I'm not getting alerts for these comments; is there a setting I'm missing?
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08/26/20

Kevin S.

tutor
Oh, I just realized you can combine the ideas. If you first substitute x-1 = ξ, then it shifts the circle to the origin, so the polar bounds are better! Somehow similar to my substitution, but also not.
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08/26/20

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