Mathematics-I (3110014)

Winter-2019

Question-5c-OR

BE | Semester-1 Winter-2019 | 17-01-2020

Q5) (c)

Evaluate $\int \int \int xyz dx dy dz$ over the positive octant of the sphere $x^{2} + y^{2} + z^{2} = 4$ .

$Let, x = r sinθ cosϕ; y = r sinθ sinϕ; z = r cosθ .$

$Jacobian, J = |\begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} & \frac{\partial x}{\partial ϕ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} & \frac{\partial y}{\partial ϕ} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial θ} & \frac{\partial z}{\partial ϕ} \end{matrix}|$

$Jacobian, J = |\begin{matrix} sinθ cosϕ & r cosθ cosϕ & - r sinθ sinϕ \\ sinθ sinϕ & r cosθ sinϕ & r sinθ cosϕ \\ cosθ & - r sinθ & 0 \end{matrix}|$

$Jacobian, J = sinθ cosϕ (0 + r^{2} \sin^{2} θ cosϕ) - r cosθ cosϕ (0 - r sinθ cosθ cosϕ) - r sinθ sinϕ (- r \sin^{2} θ sinϕ - r \cos^{2} θ sinϕ)$

$Jacobian, J = r^{2} \sin^{3} θ \cos^{2} ϕ + r^{2} sinθ \cos^{2} θ \cos^{2} ϕ + r^{2} \sin^{3} θ \sin^{2} ϕ + r^{2} sinθ \cos^{2} θ \sin^{2} ϕ$

$Jacobian, J = r^{2} \sin^{3} θ \cos^{2} ϕ + r^{2} \sin^{3} θ \sin^{2} ϕ + r^{2} sinθ \cos^{2} θ \cos^{2} ϕ + r^{2} sinθ \cos^{2} θ \sin^{2} ϕ$

$Jacobian, J = r^{2} \sin^{3} θ (\cos^{2} ϕ + \sin^{2} ϕ) + r^{2} sinθ \cos^{2} θ (\cos^{2} ϕ + \sin^{2} ϕ)$

$Jacobian, J = r^{2} \sin^{3} θ + r^{2} sinθ \cos^{2} θ$

$Jacobian, J = r^{2} sinθ (\sin^{2} θ + \cos^{2} θ)$

$Jacobian, J = r^{2} sinθ$

$Now, ∭ xyz dx dy dz$

$N, = \int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} \int_{0}^{2} (r sinθ cosϕ) (r sinθ sinϕ) (r cosθ) (r^{2} sinθ) d r d θ dϕ$

$N, = \int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} \int_{0}^{2} (r^{5}) (\sin^{3} θ cosθ) (cosϕ sinϕ) d r d θ dϕ$

$N, = (\int_{0}^{\frac{π}{2}} (cosϕ sinϕ) dϕ) (\int_{0}^{\frac{π}{2}} (\sin^{3} θ cosθ) d θ) (\int_{0}^{2} (r^{5}) d r)$

$N, = (\int_{0}^{\frac{π}{2}} (\frac{\sin 2 ϕ}{2}) dϕ) (\int_{0}^{\frac{π}{2}} (\sin^{3} θ cosθ) d θ) (\int_{0}^{2} (r^{5}) d r)$

$N, = {[- \frac{\cos 2 ϕ}{2}]}_{0}^{\frac{π}{2}} {[\frac{\sin^{4} θ}{4}]}_{0}^{\frac{π}{2}} {[\frac{r^{6}}{6}]}_{0}^{2}$

$N, = [- \frac{\cos 2 (\frac{π}{2}) - \cos 0}{2}] [\frac{\sin^{4} \frac{π}{2} - \sin^{4} 0}{4}] [\frac{2^{6} - 0^{6}}{6}]$

$N, = [- \frac{- 1 - 1}{4}] [\frac{1 - 0}{4}] [\frac{64 - 0}{6}]$

$N, = \frac{2}{4} \cdot \frac{1}{4} \cdot \frac{64}{6}$

$N, = \frac{4}{3}$

$Hence, ∭ xyz dx dy dz = \frac{4}{3}$

Mathematics-I (3110014)

Q5) (c)

Evaluate $\int \int \int xyz dx dy dz$ over the positive octant of the sphere $x^{2} + y^{2} + z^{2} = 4$ .

Questions

Go to Question Paper

Mathematics-I (3110014)

Q5) (c)

Evaluate ∫∫∫xyz dx dy dz over the positive octant of the sphere x2+y2+z2=4.

Questions

Go to Question Paper

Evaluate $\int \int \int xyz dx dy dz$ over the positive octant of the sphere $x^{2} + y^{2} + z^{2} = 4$ .