volume-bounded-by-the-curve-z-3x-2-3y-2-and-x-2-y-2-z-2-6-2-

Question Number 211749 by universe last updated on 19/Sep/24

v o l u m e b o u n d e d b y t h e c u r v e

z = \sqrt{3 x^{2} + 3 y^{2}} a n d x^{2} + y^{2} + z^{2} = 6^{2}

Answered by mr W last updated on 20/Sep/24

Commented by mr W last updated on 20/Sep/24

z = \sqrt{3 (x^{2} + y^{2})} = \sqrt{3} r

h = \sqrt{3} r

{(\sqrt{3} r)}^{2} + r^{2} = R^{2} = 6^{2}

\Rightarrow r = 3 \Rightarrow h = 3 \sqrt{3}

V = \frac{2 π R^{2} (R - h)}{3} = \frac{2 π \times 6^{2} \times 3 (2 - \sqrt{3})}{3}

= 72 (2 - \sqrt{3}) π

Commented by universe last updated on 20/Sep/24

t h a n k y o u s i r

Commented by universe last updated on 20/Sep/24

c a n u e x p l a i n V = \frac{2 π R^{2} (R - r)}{3}

Commented by mr W last updated on 20/Sep/24

Commented by universe last updated on 20/Sep/24

t h a n k s s i r

Answered by BHOOPENDRA last updated on 20/Sep/24

Answered by BHOOPENDRA last updated on 20/Sep/24

V = \int \int \int ρ^{2} s i n ϕ d ρ d θ d ϕ

= \int_{0}^{\frac{π}{6}} \int_{0}^{2 π} \int_{0}^{6} ρ^{2} s i n ϕ d ρ d θ d ϕ

= \int_{0}^{\frac{π}{6}} \int_{0}^{2 π} (\frac{6^{3} - 0^{3}}{3}) s i n ϕ d θ d ϕ

= \int_{0}^{\frac{π}{6}} 72 (2 π - 0) s i n ϕ d ϕ

= - 144 π c o s ϕ ∣_{0}^{\frac{π}{6}}

= 144 π (\frac{2 - \sqrt{3}}{2})

= 72 π (2 - \sqrt{3})

Commented by universe last updated on 20/Sep/24

t h a n k y o u s i r

Commented by mr W last updated on 20/Sep/24

V = 72 π (2 - \sqrt{3}) > 0

Commented by BHOOPENDRA last updated on 20/Sep/24

T h a n k s M r . W i t w a s t y p o