Integrate the scalar field f(\mathbf{r}) = y over the area that is enclose by the positive cartesian coordinate axes and the parabola y=1-x^2 . The answer is

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We are integrating f(r) = y over the region in the first quadrant bounded by the positive x- and y-axes and the parabola y = 1 - x^2.
First, identify the horizontal extent of the region: along x from 0 to where y = 0 on the parabola, which gives 1 - x^2 = 0, so x = 1 in the first quadrant.
For a fixed x between 0 and 1, y runs from 0 up to y = 1 - x^2, since the region lies under the parabola and above the x-axis.
Set up the double integral: ∫_{x=0}^{1} ∫_{y=0}^{1-x^2} y dy dx.
Evaluate the inner integral with respect to y: ∫_{0}^{1-x^2} y dy = [ (1/2) y^2 ]_{0}^{1-x^2} = (1/2)(1 - x^2)^2.
Thus the problem reduces to a single integral: ∫_{0}^{1} (1/2)(1 - x^2)^2 dx = (1/2) ∫_{0}^{1} (1 - 2x^2 + x^4) dx.
Compute term by term: ∫_{0}^{1} 1 dx = 1, ∫_{0}^{1} x^2 dx = 1/3, ∫_{0}^{1} x^4 dx = 1/5.
Combine: (1/2)[ 1 - 2(1/3) + (1/5) ] = (1/2)[ 1 - 2/3 + 1/5 ].
Find a common denominator, say 15: 1 = 15/15, 2/3 = 10/15, 1/5 = 3/15, so inside the brackets we have (15/15 - 10/15 + 3/15) = 8/15.
Multiply by 1/2: (1/2) * (8/15) = 4/15.
Therefore, the value of the integral over the specified region is 4/15.

PHAS0009_24-25 2.3 Quiz 2 [0.25 hrs]

Integrate the scalar field f(\mathbf{r}) = y over the area that is enclose by the positive cartesian coordinate axes and the parabola y=1-x^2 . The answer is

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