In class Problem

asked 2014-07-23 09:41:37 -0600

Anonymous
740 ●2 ●27 ●49

In class today my group and I had a little problem solving the second problem written on the board: Evaluate $\int\int\int sin((x^2+y^2+z^2)^{3/2}) dV$ where $D$ is defined as the top half of the solid unit sphere.

This is how far we got $$\int_0^\pi\int_0^\pi\int_0^1 (sin((p^2)^{3/2})p^2 sin\phi dP d\phi d\Theta$$ $$\pi\int_0^\pi\int_0^1 sin(p^3)p^2 sin\phi dP d\phi$$ $$u=p^3$$ $$1/3 du=p^2 dP$$ $$\frac{\pi}{3}\int_0^\pi\int_0^1 sin(u) sin\phi du d\phi$$ $$\frac{\pi}{3}\int_0^\pi sin\phi - cos(u) \biggr|_0^1 d\phi$$ $$\frac{\pi}{3}\int_0^\pi 1-cos(1) d\phi$$ $$\frac{\pi}{3} (\phi-\phi cos(1)) \biggr|_0^\pi$$ $$\frac{\pi}{3} (\pi - \pi cos(1) $$

Did we start with the correct domain of integration or did we make a mistake in the calculations?

answered 2014-07-23 11:45:41 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

The way our group set it up was different, and here is why.

The domain is only the top half of the unit sphere, so $\delta\phi$ is only going from zero to $\frac{\pi}{2}$

Since it is all the way around the middle of the sphere, for lack of better wording, $\delta\theta$ will go from zero to $2\pi$.

Giving us the following:

$$\int_0^{2\pi}\int_0^\frac{\pi}{2}\int_0^1 (\sin((\rho^2)^{3/2})\rho^2 \sin\phi d\rho d\phi d\Theta$$ $${2\pi}\int_0^\frac{\pi}{2}\int_0^1 \sin(\rho^3)\rho^2 \sin\phi d\rho d\phi$$

Using u sub just like you did... $$u=\rho^3$$ $$1/3 du=\rho^2 d\rho$$ $$\frac{{2\pi}}{3}\int_0^\frac{\pi}{2}\int_0^1 \sin(u) \sin\phi du d\phi$$ $$\frac{{2\pi}}{3}\int_0^\frac{\pi}{2} \sin\phi( - \cos(u)) \biggr|_0^1 d\phi$$ $$\frac{{2\pi}}{3}\int_0^\frac{\pi}{2} \sin\phi(-\cos(1)+1) d\phi$$

$$\frac{{2\pi}}{3}(-\cos(1)+1) (-\cos(\phi)) \biggr|_0^\frac{\pi}{2}$$ $$\frac{{2\pi}}{3}(-\cos(1)+1) (1) $$ $$\frac{{2\pi}}{3}(1-\cos(1)) $$

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In class Problem

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