How Accurate Is the Substitution in This Double Integral Problem?

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Homework Help Overview

The discussion revolves around a double integral problem involving the substitution of variables and the conversion to polar coordinates. The original poster presents an integral with a specific area of integration defined by an elliptical region.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts a variable substitution and expresses concern about potential errors in their approach. Some participants question the validity of the substitution and the subsequent polar coordinate transformation, suggesting that the area of integration is misrepresented. Others raise issues regarding the limits of integration, particularly for the radial component in polar coordinates.

Discussion Status

Participants are actively engaging with the original poster's approach, providing feedback on the correctness of the substitutions and limits. There is a recognition of a typo in the original statement, and some participants have offered clarifications regarding the limits of integration, particularly for the radius in polar coordinates.

Contextual Notes

There is a noted correction regarding the area of integration, which is defined by an elliptical region rather than a circular one. The discussion reflects a collaborative effort to clarify the setup and assumptions of the problem.

Swasse
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Hey.

Homework Statement



∫∫x^3 dxdy, with the area of integration: D={(x,y)∈R^2: 1<=x^2+9y^2<=9, x>=3y}

The Attempt at a Solution



Did the variable substitution u=x and v=3y so the area of integration became 1<=u^2 + v^2 <=9, u>=v. And the integral became ∫∫(1/3)u^3 dudv. Then I switched to polar coordinates and the integral ∫∫(1/3)r^4cos^3θ drdθ, with 1<=r<=9 and -3/4<=θ<=pi/4.

Any errors so far?

I really appreciate any help you can provide.

Homework Statement


Homework Equations


The Attempt at a Solution

 
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Well, first, if you take u= x, v= 3y then 1<= x^2+ y^2<= 9 does NOT become 1<= u^2+ v^2<= 9, it becomes 1<= u^2+ v^2/9<= 9, the region between two ellipses. Since you then change to polar coordinates, I don't see any good reason to do that. Rather, change to polar coordinates initially. The line x= 3y, or y= x/3 has \theta= arctan(1/3) so the integral is \int_{\theta= -\pi+ arctan(1/3)}^{arctan(1/3)}\int_{r= 1}^3 r^4 cos^3(\theta) drd\theta.
 
HallsofIvy said:
Well, first, if you take u= x, v= 3y then 1<= x^2+ y^2<= 9 does NOT become 1<= u^2+ v^2<= 9, it becomes 1<= u^2+ v^2/9<= 9, the region between two ellipses. Since you then change to polar coordinates, I don't see any good reason to do that. Rather, change to polar coordinates initially. The line x= 3y, or y= x/3 has \theta= arctan(1/3) so the integral is \int_{\theta= -\pi+ arctan(1/3)}^{arctan(1/3)}\int_{r= 1}^3 r^4 cos^3(\theta) drd\theta.

My mistake, it should have said 1<= x^2+ 9y^2<= 9
 
Can someone please confirm if my way of solving is correct? I made a typo in the first post, but I have corrected it now.
 
Your upper limit for ##r## is wrong.
 
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vela said:
Your upper limit for ##r## is wrong.

I had a feeling it was, because my answer was in the right form, just too large. Can you please explain why it is wrong?
 
vela said:
Your upper limit for ##r## is wrong.

Just figured it out, the radius is not 9, it is √9=3. Is that the correct limit?
 
Yup!
 
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