In evaluation of the iterative integral $\int_{0}^{1}\int_{0}^{1}\frac{y}{(1+x^2+y^2)^{3/2}}d yd x,$ I want to show the details of the whole calculation, and so, the question comes out, how to express the partial differential of $1+x^2+y^2$ with respect to just $y,$ not to $x?$ At present, I write it as $d_y(1+x^2+y^2),$ where the subscript $y$ denotes the partial differential is to $y,$ not $x.$ See below: \begin{align*} &\int_{0}^{1}\int_{0}^{1}\frac{y}{(1+x^2+y^2)^{3/2}}d yd x =\frac{1}{2}\int_{0}^{1}\int_{0}^{1}(1+x^2+y^2)^{-3/2}d(y^2) d x \\ =&\frac{1}{2}\int_{0}^{1}\int_{0}^{1}(1+x^2+y^2)^{-3/2}d_y(1+x^2+y^2) dx=\frac{1}{2}\int_{0}^{1}(-2)(1+x^2+y^2)^{-1/2}\Big|_{y=0}^{y=1}d x\\ =&- \int_{0}^{1}\big((2+x^2)^{-1/2}-(1+x^2)^{-1/2}\big)d x= \int_{0}^{1}\left(\frac{1}{\sqrt{1+x^2}}-\frac{1}{\sqrt{2+x^2}}\right)d x\\ =&\left(\ln\big(x+\sqrt{1+x^2}\,\big)-\ln\big(x+\sqrt{2+x^2}\,\big)\right)\Big|_0^1=\ln\frac{\sqrt{2}+2}{1+\sqrt{3}}. \end{align*} Is there standard notation for the partial differential of $1+x^2+y^2$ with respect to $y?$
2026-03-25 04:54:41.1774414481
Notation for partial differential in evaluation of iterative integrals
23 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in MULTIPLE-INTEGRAL
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