- #1
jdstokes
- 520
- 1
Hi all,
Let D be the top half of a ball of radius a>0 and let
\mathbf{f} = xz \mathbf{i} + y \mathbf{j} + x \mathbf{k}
Calculate the outward flux across S using the definition of the surface integral.
\int_{S} \mathbf{f} \cdot \mathbf{n} dS = \int_{cap} \mathbf{f} \cdot \mathbf{n} dS + \int_{disc} \mathbf{f} \cdot \mathbf{n} dS
\frac{1}{a}\int_{cap} (xz,y,x) \cdot (x,y,z) d S
\frac{1}{a}\int_{cap} x^2z + y^2 + xz d S
\frac{1}{a}\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^2\theta\cos\theta + r^2\sin^2\varphi\sin^2\theta + r^2\cos\varphi\sin\theta\cos\theta) (a^2 \sin\theta) d\theta d\varphi
a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\sin^2\varphi\sin^3\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi
a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi
I assume that one would then evaluate the integral by parts. But is there an easier method?
Let D be the top half of a ball of radius a>0 and let
\mathbf{f} = xz \mathbf{i} + y \mathbf{j} + x \mathbf{k}
Calculate the outward flux across S using the definition of the surface integral.
\int_{S} \mathbf{f} \cdot \mathbf{n} dS = \int_{cap} \mathbf{f} \cdot \mathbf{n} dS + \int_{disc} \mathbf{f} \cdot \mathbf{n} dS
\frac{1}{a}\int_{cap} (xz,y,x) \cdot (x,y,z) d S
\frac{1}{a}\int_{cap} x^2z + y^2 + xz d S
\frac{1}{a}\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^2\theta\cos\theta + r^2\sin^2\varphi\sin^2\theta + r^2\cos\varphi\sin\theta\cos\theta) (a^2 \sin\theta) d\theta d\varphi
a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\sin^2\varphi\sin^3\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi
a\int_{0}^{2\pi}\int_{0}^{\pi/2} (r^3\cos^2\varphi\sin^3\theta\cos\theta + r^2\cos\varphi\sin^2\theta\cos\theta) d\theta d\varphi
I assume that one would then evaluate the integral by parts. But is there an easier method?
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