Reverse Integration: Solving \int_0^3\int_{y^2}^9 y \cos{x^2} dxdy

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Homework Help Overview

The discussion revolves around evaluating a double integral by reversing the order of integration. The integral in question is \(\int_0^3\int_{y^2}^9 y \cos{x^2} dxdy\), which involves concepts from calculus related to integration and area under curves.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the limits of integration after reversing the order, with one participant suggesting limits of \(\sqrt{x} \leq y \leq 3\) and \(0 \leq x \leq 9\), while another proposes \(0 \leq y \leq \sqrt{x}\) and \(0 \leq x \leq 9\). There is also a mention of verifying the area through both integrals.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the limits of integration. Some guidance has been offered regarding the visualization of the area, but no consensus has been reached on the correct limits.

Contextual Notes

Participants are working under the constraints of reversing the order of integration and ensuring the limits accurately represent the area of integration. There is an emphasis on visualizing the region defined by the original limits.

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Homework Statement



Evaluate the integral by reversing the order of integration.

Homework Equations



\int_0^3\int_{y^2}^9 y \cos{x^2} dxdy

The Attempt at a Solution



I want to make sure I'm right so far before going on. I have the reversed limits as:
\sqrt{x} <= y <= 3
0 <= x <= 9

Is this right?
 
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i would try drawing the area, could be wrong, but i think it looks more like:
0 <= y <= \sqrt{x}
0 <= x <= 9
 
you could also try both integrals to check you get the same area, though i suppose that's why you;re reversing the order in the first place
 
Thanks!
 

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