With the help of triple integrals, find the volume of the sphere ρ = 2 in a) Rectangular coordianates b) Cylinderical coordiantes c) Spherical coordinates
Solution:
a) Rectangular coordinates:
x 2 + y 2 + z 2 = 4 x^2+y^2+z^2=4 x 2 + y 2 + z 2 = 4
V = ∫ − 2 2 d x ∫ − 4 − x 2 4 − x 2 d y ∫ − 4 − x 2 − y 2 4 − x 2 − y 2 d z = V=\displaystyle\int_{-2}^2dx\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}dz= V = ∫ − 2 2 d x ∫ − 4 − x 2 4 − x 2 d y ∫ − 4 − x 2 − y 2 4 − x 2 − y 2 d z = 8 ∫ 0 2 d x ∫ 0 4 − x 2 d y ∫ 0 4 − x 2 − y 2 d z = 8\displaystyle\int_{0}^2dx\int_{0}^{\sqrt{4-x^2}}dy\int_{0}^{\sqrt{4-x^2-y^2}}dz= 8 ∫ 0 2 d x ∫ 0 4 − x 2 d y ∫ 0 4 − x 2 − y 2 d z = 8 ∫ 0 2 d x ∫ 0 4 − x 2 4 − x 2 − y 2 d y 8\displaystyle\int_{0}^2dx\int_{0}^{\sqrt{4-x^2}}\sqrt{4-x^2-y^2}dy 8 ∫ 0 2 d x ∫ 0 4 − x 2 4 − x 2 − y 2 d y
Let's apply the change of variables:
y = 4 − x 2 ⋅ sin t y=\sqrt{4-x^2}\cdot\sin{t} y = 4 − x 2 ⋅ sin t
d y = 4 − x 2 ⋅ cos t ⋅ d t dy=\sqrt{4-x^2}\cdot\cos{t}\cdot dt d y = 4 − x 2 ⋅ cos t ⋅ d t
y = 0 ⟹ t = 0 y=0 \implies t=0 y = 0 ⟹ t = 0
y = 4 − x 2 ⟹ t = π 2 y=\sqrt{4-x^2}\implies t=\frac{\pi}{2} y = 4 − x 2 ⟹ t = 2 π
V = 8 ∫ 0 2 d x ∫ 0 π / 2 4 − x 2 − ( 4 − x 2 ) sin 2 t ⋅ V=8\displaystyle\int_{0}^2dx\int_{0}^{\pi/2}\sqrt{4-x^2-(4-x^2)\sin^2{t}}\cdot V = 8 ∫ 0 2 d x ∫ 0 π /2 4 − x 2 − ( 4 − x 2 ) sin 2 t ⋅ 4 − x 2 cos t d t = \sqrt{4-x^2}\cos{t}dt= 4 − x 2 cos t d t =
8 ∫ 0 2 d x ∫ 0 π / 2 ( 4 − x 2 ) cos 2 t d t = 8\displaystyle\int_{0}^2dx\int_{0}^{\pi/2}(4-x^2)\cos^2{t}dt= 8 ∫ 0 2 d x ∫ 0 π /2 ( 4 − x 2 ) cos 2 t d t = 8 ∫ 0 2 d x ∫ 0 π / 2 ( 4 − x 2 ) ( 1 + cos 2 t ) 2 d t = 8\displaystyle\int_{0}^2dx\int_{0}^{\pi/2}(4-x^2)\frac{(1+\cos{2t})}{2}dt= 8 ∫ 0 2 d x ∫ 0 π /2 ( 4 − x 2 ) 2 ( 1 + cos 2 t ) d t = 8 ∫ 0 2 d x ( 4 − x 2 ) ( t + sin 2 t 2 ) 2 ∣ 0 π / 2 = 8\displaystyle\int_{0}^2dx(4-x^2)\frac{(t+\frac{\sin{2t}}{2})}{2}|_0^{\pi/2}= 8 ∫ 0 2 d x ( 4 − x 2 ) 2 ( t + 2 s i n 2 t ) ∣ 0 π /2 = 2 π ∫ 0 2 ( 4 − x 2 ) d x = 2 π ( 4 x − x 3 / 3 ) ∣ 0 2 = 32 π 3 2\pi\displaystyle\int_{0}^2(4-x^2)dx=2\pi(4x-x^3/3)|_0^2=\frac{32\pi}{3} 2 π ∫ 0 2 ( 4 − x 2 ) d x = 2 π ( 4 x − x 3 /3 ) ∣ 0 2 = 3 32 π
b) Cylinderical coordiantes:
x = ρ cos φ x=\rho\cos{\varphi} x = ρ cos φ
y = ρ sin φ y=\rho\sin{\varphi} y = ρ sin φ
z = z z=z z = z
J = ρ J=\rho J = ρ
The equation of the sphere:
ρ 2 + z 2 = 4 \rho^2+z^2=4 ρ 2 + z 2 = 4
V = ∫ 0 2 π d φ ∫ 0 2 d ρ ∫ − 4 − ρ 2 4 − ρ 2 ρ d z = V=\displaystyle\int_0^{2\pi}d\varphi\int_0^2d\rho\int_{-\sqrt{4-\rho^2}}^{\sqrt{4-\rho^2}}\rho dz= V = ∫ 0 2 π d φ ∫ 0 2 d ρ ∫ − 4 − ρ 2 4 − ρ 2 ρ d z = 2 ∫ 0 2 π d φ ∫ 0 2 d ρ ∫ 0 4 − ρ 2 ρ d z = 2\displaystyle\int_0^{2\pi}d\varphi\int_0^2d\rho\int_0^{\sqrt{4-\rho^2}}\rho dz= 2 ∫ 0 2 π d φ ∫ 0 2 d ρ ∫ 0 4 − ρ 2 ρ d z = 4 π ∫ 0 2 4 − ρ 2 ρ d ρ = 4\pi\displaystyle\int_0^2\sqrt{4-\rho^2}\rho d\rho= 4 π ∫ 0 2 4 − ρ 2 ρ d ρ = 2 π ∫ 0 2 4 − ρ 2 d ρ 2 = − 4 π 3 ( 4 − ρ 2 ) 3 / 2 ∣ 0 2 = 32 π 3 2\pi\displaystyle\int_0^2\sqrt{4-\rho^2} d\rho^2=-\frac{4\pi}{3}\displaystyle(4-\rho^2)^{3/2}|_0^2=\frac{32\pi}{3} 2 π ∫ 0 2 4 − ρ 2 d ρ 2 = − 3 4 π ( 4 − ρ 2 ) 3/2 ∣ 0 2 = 3 32 π
c) Spherical coordinates
x = r sin θ cos φ x=r\sin{\theta}\cos{\varphi} x = r sin θ cos φ
y = r sin θ sin φ y=r\sin{\theta}\sin{\varphi} y = r sin θ sin φ
z = r cos θ z=r\cos{\theta} z = r cos θ
J = r 2 sin θ J=r^2\sin{\theta} J = r 2 sin θ
V = ∫ 0 2 π d φ ∫ 0 π d θ ∫ 0 2 r 2 sin θ d r = V=\displaystyle\int_0^{2\pi}d\varphi\int_0^{\pi}d\theta\int_0^2r^2\sin{\theta}dr= V = ∫ 0 2 π d φ ∫ 0 π d θ ∫ 0 2 r 2 sin θ d r = 2 π ∫ 0 π d θ sin θ r 3 3 ∣ 0 2 = 2\pi\displaystyle\int_0^{\pi}d\theta\sin{\theta}\frac{r^3}{3}|_0^2= 2 π ∫ 0 π d θ sin θ 3 r 3 ∣ 0 2 = 16 π 3 ∫ 0 π sin θ d θ = \displaystyle\frac{16\pi}{3}\int_0^{\pi}\sin{\theta}d\theta= 3 16 π ∫ 0 π sin θ d θ = − 16 π 3 cos θ ∣ 0 π = 32 π 3 -\displaystyle\frac{16\pi}{3}\cos{\theta}|_0^{\pi}=\frac{32\pi}{3} − 3 16 π cos θ ∣ 0 π = 3 32 π
Answer: V = 32 π 3 . V=\displaystyle\frac{32\pi}{3}. V = 3 32 π .
#339914 on Dec 2023
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