Convert ∫04∫016−y2∫x2+y232−x2−y2(x2+y2+z2)dzdxdy into spherical coordinates. ∫0π/4∫0π/2∫042ρ4sinϕdρdθdϕ∫0π/4∫0π/2∫04ρ4sinϕdρdθdϕ∫0π/4∫−π/2π/2∫042ρ4sinϕdρdθdϕ∫0π/3∫0π/2∫042ρ4sinϕdρdθdϕ∫0π/4∫0π/2∫042ρ4sinϕdρdϕdθ∫0π/4∫−π/2π/2∫042ρ4cosϕdρdθdϕ∫02π/3∫−π/2π/2∫042ρ4cosϕdρdϕdθ

Given that the integral is int_0^4 int_0^(sqrt(16-y^2)) int_(sqrt(x^2+y^2))^(sqrt(32-x^2-y^2)) (x^2+y^2+z^2) dzdxdy We want to convert into spherical coordinates

Resuelto Convert ∫04∫016−y2∫x2+y232−x2−y2(x2+y2+z2)dzdxdy into

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Pregunta: Convert ∫04∫016−y2∫x2+y232−x2−y2(x2+y2+z2)dzdxdy into spherical coordinates. ∫0π/4∫0π/2∫042ρ4sinϕdρdθdϕ∫0π/4∫0π/2∫04ρ4sinϕdρdθdϕ∫0π/4∫−π/2π/2∫042ρ4sinϕdρdθdϕ∫0π/3∫0π/2∫042ρ4sinϕdρdθdϕ∫0π/4∫0π/2∫042ρ4sinϕdρdϕdθ∫0π/4∫−π/2π/2∫042ρ4cosϕdρdθdϕ∫02π/3∫−π/2π/2∫042ρ4cosϕdρdϕdθ
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Solución
Paso 1
Given that the integral is
$\int_{0}^{4} \int_{0}^{\sqrt{16 - y^{2}}} \int_{\sqrt{x^{2} + y^{2}}}^{\sqrt{32 - x^{2} - y^{2}}} (x^{2} + y^{2} + z^{2})$ $d z d x d y$
We want to convert into spherical coordinates
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Paso 2
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Paso 3
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Respuesta
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Texto de la transcripción de la imagen:

Convert

\int_{0}^{4} \int_{0}^{16 - y^{2}} \int_{x^{2} + y^{2}}^{32 - x^{2} - y^{2}} (x^{2} + y^{2} + z^{2}) d z d x d y

into spherical coordinates.

\int_{0}^{π /4} \int_{0}^{π /2} \int_{0}^{42} ρ^{4} sin ϕ d ρ d θ d ϕ \int_{0}^{π /4} \int_{0}^{π /2} \int_{0}^{4} ρ^{4} sin ϕ d ρ d θ d ϕ \int_{0}^{π /4} \int_{- π /2}^{π /2} \int_{0}^{42} ρ^{4} sin ϕ d ρ d θ d ϕ \int_{0}^{π /3} \int_{0}^{π /2} \int_{0}^{42} ρ^{4} sin ϕ d ρ d θ d ϕ \int_{0}^{π /4} \int_{0}^{π /2} \int_{0}^{42} ρ^{4} sin ϕ d ρ d ϕ d θ \int_{0}^{π /4} \int_{- π /2}^{π /2} \int_{0}^{42} ρ^{4} cos ϕ d ρ d θ d ϕ \int_{0}^{2 π /3} \int_{- π /2}^{π /2} \int_{0}^{42} ρ^{4} cos ϕ d ρ d ϕ d θ

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