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Multiply the first equation by ", StyleBox["2", "Input"], " and add it to the second:\n", StyleBox[" 2u == 4x^2 - 2y^2", "Input"], StyleBox["\n", "Textt"], StyleBox[ " v == 2y^2 - x^2\n 2u + v == 3x^2\n x == Sqrt[(2u + v)/3]\n", "Input"], "Similarly,\n", StyleBox[" y == Sqrt[(2v + u)/3]\n", "Input"], "We take the positive roots since the region is in the first quadrant. 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J[f,g,h,{u,v,w}] gives the Jacobian \ determinant of the coordinate transformation x = f(u,v,w), y = g(u,v,w), z = \ h(u,v,w)."\)], "Print"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "J[", StyleBox["Sqrt[(2u + v)/3]", "Input"], ",", StyleBox["Sqrt[(2v + u)/3]", "Input"], ",{u,v}]" }], "Input"], Cell[BoxData[ \(1\/\(4\ \@\(2\ u + v\)\ \@\(u + 2\ v\)\)\)], "Output"] }, Open ]], Cell[TextData[{ "The function in the new coordinates is ", StyleBox["x y -> Sqrt[(2u + v)/3]", "Input"], " ", StyleBox["Sqrt[(2v + u)/3]", "Input"], ". Multiplying by the Jacobian gives:" }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Sqrt[(2u + v)/3]", "Input"], " ", StyleBox["Sqrt[(2v + u)/3] * j", "Input"] }], "Input"], Cell[BoxData[ \(1\/12\)], "Output"] }, Open ]], Cell[TextData[{ "The new region is, of course, described by\n", StyleBox[" 1 <= u <= 2\n 1 <= v <= 2\n", "Input"], "so the integral of ", StyleBox["f[x,y]", "Input"], " over ", StyleBox["R", "Input"], " equals" }], "Text"], Cell[CellGroupData[{ Cell["Integrate[1/12, {u,1,2}, {v,1,2}]", "Input"], Cell[BoxData[ \(1\/12\)], "Output"] }, Open ]], Cell[TextData[{ "It is worth noting that this problem can also be done easily by taking \ advantage of the formula:\n", StyleBox[ " J[x[u,v], y[u,v], {u,v}] == 1/J[u[x,y], v[x,y], {x,y}]\n", "Input"], "This formula is not always \"useful\" because the right hand side is still \ in terms of the variables ", StyleBox["x", "Input"], " and ", StyleBox["y", "Input"], ". In this problem however:" }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "m = {{D[", StyleBox["2x^2-y^2", "Input"], ",x],D[", StyleBox["2x^2-y^2", "Input"], ",y]},\n {D[", StyleBox["2y^2-x^2", "Input"], ",x],D[", StyleBox["2y^2-x^2", "Input"], ",y]}};\nm //MatrixForm" }], "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(4\ x\), \(\(-2\)\ y\)}, {\(\(-2\)\ x\), \(4\ y\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["j = Det[m]", "Input"], Cell[BoxData[ \(12\ x\ y\)], "Output"] }, Open ]], Cell[TextData[{ "Thus, the new integrand is ", StyleBox["x y (1/j) == 1/12", "Input"], " as above", "!" }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Let ", StyleBox["C", "Input"], StyleBox[" be the curve ", "Text"], StyleBox["y == x^2", "Input"], StyleBox[" from ", "Text"], StyleBox["0 <= x <= 1", "Input"], StyleBox[". Evaluate the line integral of ", "Text"], StyleBox["f[x_,y_] = Sqrt[y]", "Input"], StyleBox[" over the curve ", "Text"], StyleBox["C.", "Input"] }], "Subsubsection", CellMargins->{{Inherited, 112}, {Inherited, Inherited}}, Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Solution"], "Subsubsection", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell["The curve is parameterized by", "Text"], Cell["r[t_] = {t, t^2};", "Input"], Cell[TextData[{ "for ", StyleBox["0 <= t <= 1", "Input"], ". 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