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Calculus Level 5

2 2 4 x 2 4 x 2 4 x 2 y 2 4 x 2 y 2 ( x 2 + y 2 + z 2 ) 3 2 d x d y d z \large\displaystyle\int \limits^{2}_{-2} \displaystyle\int \limits^{\sqrt{4- x^{2}} }_{-\sqrt{4-x^{2}} }\displaystyle\int \limits^{\sqrt{4- x^{2}- y^{2}} }_{-\sqrt{4-x^{2}- y^{2}} }\left( x^{2}+ y^{2}+ z^{2}\right) ^{\frac{3}{2} }\, dx\, dy \, dz

If the value of triple integral above equals to A B π \frac AB \pi for coprime positive integers A A and B B , find the value of A + B A+B .

Image Credit: Flickr lalique7 .


The answer is 131.

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1 solution

Refaat M. Sayed
Jul 30, 2015

At first chaneg to spheric coordinates x = r c o s ( ϕ ) s i n ( θ ) x = rcos ( \phi)sin (\theta) y = r s i n ( ϕ ) s i n ( θ ) y = r sin (\phi)sin (\theta) z = r c o s ( θ ) z = r cos(\theta) d x d y d z = r 2 s i n ( θ ) d r d ϕ d θ dx dy dz = r^{2} sin(\theta) dr d\phi d\theta 0 r 4 c o s ( θ ) 0 \leq r \leq 4cos(\theta) 0 ϕ 2 π 0\leq \phi \leq 2\pi 0 θ π 2 0\leq \theta \leq\frac{\pi}{2} I = 0 2 π 0 π 2 0 4 c o s ( θ ) r 3 ( r 2 s i n ( θ ) ) d ϕ d θ d r I=\displaystyle\int \limits^{2\pi }_{0}\displaystyle\int \limits^{\frac{\pi }{2} }_{0}\displaystyle\int \limits^{4cos\left( \theta \right) }_{0}r^{3}\left( r^{2}sin\left( \theta \right) \right) d\phi d\theta dr I = 0 2 π d ϕ 0 π 2 s i n ( θ ) d θ 0 4 c o s ( θ ) r 5 d r I=\displaystyle\int \limits^{2\pi }_{0}d\phi \displaystyle\int \limits^{\frac{\pi }{2} }_{0}sin\left( \theta \right) d\theta \displaystyle\int \limits^{4cos\left( \theta \right) }_{0}r^{5}dr I = 2 π 0 π 2 s i n ( θ ) ( 4 6 c o s 6 ( θ ) 6 ) d θ I=2\pi\displaystyle\int \limits^{\frac{\pi }{2} }_{0}sin\left( \theta \right) \left( \frac{4^{6}cos^{6}\left( \theta \right) }{6} \right) d\theta I = 4096 3 × π 7 I=\frac {4096}{3} \times \frac{\pi}{7} I = 4096 π 21 I = \frac{4096\pi}{21} so A + B = 4117 \text{so} A+B = 4117

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