Sunday, February 10, 2013

Temp Post - An Integration

Consider $a$ as a constant. \begin{align*} &2 \pi \int_1^2 (2a - ax) \sqrt{1+a^2} dx \\ =&2 \pi a \sqrt{1+a^2} ( 2 \int_1^2 1 dx - \int_1^2 x dx ) \\ =&2 \pi a \sqrt{1+a^2} \big( 2 (x|_1^2) - (\frac{1}{2}x^2|_1^2) \big) \\ =&2 \pi a \sqrt{1+a^2} (2 - (\frac{1}{2} \times 4 - \frac{1}{2})) \\ =&2 \pi a \sqrt{1+a^2} (\frac{1}{2}) \\ =&\pi a \sqrt{1+a^2} \end{align*}

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