[คุณชายน้อย]

ข้อความเดิมเขียนโดยคุณ คุณชายน้อย

ข้อ 1. $Surface=\int_{0}^{5/2} \int_{0}^{\sqrt{25-4x^2} }\frac{5}{\sqrt{25-x^2-y^2} } \,dy \,dx = \frac{25\pi}{6} $
ข้อ 2. $Surface=4\int_{0}^{\pi/2} \int_{0}^{a~sin~\theta }\frac{a}{\sqrt{a^2-r^2} } \,dr \,d\theta =\frac{a\pi^2}{2} $

ข้อความเดิมเขียนโดยคุณ gnopy

ตัวแรกอินทิเกรตออกมายังไงครับ ข้อ 1อะครับ

$Surface=\int_{0}^{5/2} \int_{0}^{\sqrt{25-4x^2} }\frac{5}{\sqrt{25-x^2-y^2} } \,dy \,dx $
$~~~~~=5\int_{0}^{5/2} arcsin\frac{y}{ \sqrt{25-4x^2} } \left|\,\right. \begin{array}{rcl} _{ \sqrt{25-4x^2} } \\ _{ y=0 } \end{array} dx $
$~~~~~=5\int_{0}^{5/2} arcsin\sqrt{\frac{25-4x^2}{25-x^2} } dx $
$~~~~~ โดยการ~ By~Part~,~u=arcsin\sqrt{\frac{25-4x^2}{25-x^2} }~ ,~ dv = dx$
$~~~~~=5 \left[\,\right. x arcsin\sqrt{\frac{25-4x^2}{25-x^2} } \left|\,\right. \begin{array}{rcl} _{ 5/2 } \\ _{ y=0 } \end{array} - \int_{0}^{5/2} \frac{25\sqrt{3}x}{(x^2-25) \sqrt{25-4x^2} } dx \left.\,\right] $
$~~~~~เริ่มเป็น~Improper~Integral$
$~~~~~=5(-25\sqrt{3} ) \int_{0}^{5/2} \frac{x}{(x^2-25) \sqrt{25-4x^2} } dx $
$~~~~~=5(\frac{-25\sqrt{3} }{2} ) \left[\,\right. \int_{0}^{5/2} \frac{1}{(x-5) \sqrt{25-4x^2} } dx + \int_{0}^{5/2} \frac{1}{(x+5) \sqrt{25-4x^2} } dx \left.\,\right] $
$~~~~~=\frac{25}{2} \left[\,\right. \int_{0}^{5/2} \frac{-5\sqrt{3} }{(x-5) \sqrt{25-4x^2} } dx - \int_{0}^{5/2} \frac{5\sqrt{3} }{(x+5) \sqrt{25-4x^2} } dx \left.\,\right] $
$~~~~~=\frac{25}{2} \left[\,\right. \int_{0}^{5/2} d(arctan\frac{4x-5}{\sqrt{3}\sqrt{25-4x^2} } ) - \int_{0}^{5/2} d(arctan\frac{4x+5}{\sqrt{3}\sqrt{25-4x^2} } ) \left.\,\right] $
$~~~~~=\frac{25}{2} \lim_{a \to {(5/2)}^{-}} \left[\,\right. arctan\frac{4x-5}{\sqrt{3}\sqrt{25-4x^2} } - arctan\frac{4x+5}{\sqrt{3}\sqrt{25-4x^2} } \left.\,\right] \begin{array}{rcl} _{ a } \\ _{ 0 } \end{array} $
$~~~~~= \frac{25}{2} \left[\,\right. ( \frac{\pi}{2}- \frac{\pi}{2} ) - ( arctan(-1/\sqrt{3})-arctan(1\sqrt{3} ) ) \left.\,\right] $
$~~~~~= \frac{25}{2} \left[\,\right. -(\frac{-\pi}{6})+\frac{\pi}{6} \left.\,\right] $
$~~~~~= \frac{25\pi}{6}$