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Use the given transformation to evaluate the integral.

∬R³x²dA, where R is the region bounded by the ellipse 25x²+16y²=400.
x=4u, y=5v
∬R 3x²dA=

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Final answer:

To evaluate the integral of 3x² over the region bounded by the given ellipse, a transformation to u and v is used to change the ellipse to a unit circle and the area element dA to 20dudv. The Jacobian of the transformation is calculated and used to express the transformed integral, which is then integrated over the unit circle to obtain the result.

Step-by-step explanation:

To evaluate the integral ∫∫ R 3x² dA, where R is bounded by the ellipse 25x²+16y²=400 using the transformation x=4u and y=5v, we first rewrite the ellipse equation in terms of u and v. After substitution, the equation becomes u²+v²=1, representing a unit circle.

Next, we need to determine the Jacobian of the transformation, which is the determinant of the matrix of partial derivatives of the transformation. The result is J = |dx/du dy/dv| = |4 0| |0 5| = 20. Therefore, the area element dA is transformed to 20dudv.

The integral becomes ∫∫_R 3x² dA = ∫∫_R 3(4u)² · 20dudv = ∫∫_R 960u² dudv

Integrate over the unit circle: ∫∫_R 960u² dudv = 960 ∫∫_{unit circle} u² dudv

The calculation is straightforward due to the symmetry of the problem. Finally, by integrating over the unit circle, we get the result.

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