Chapter 13: Problem 4
Explain why the element of area in Cartesian coordinates \(d x d y\) becomes \(r d r d \theta\) in polar coordinates.
Chapter 13: Problem 4
Explain why the element of area in Cartesian coordinates \(d x d y\) becomes \(r d r d \theta\) in polar coordinates.
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Use a change of variables to evaluate the following integrals. $$\begin{aligned} &\iiint_{D} x y d V ; D \text { is bounded by the planes } y-x=0\\\ &y-x=2, z-y=0, z-y=1, z=0, \text { and } z=3 \end{aligned}$$
In Section 10.3 it was shown that the area of a region enclosed by the polar curve \(r=g(\theta)\) and the rays \(\theta=\alpha\) and \(\theta=\beta,\) where \(\beta-\alpha \leq 2 \pi,\) is \(A=\frac{1}{2} \int_{\alpha}^{\beta} r^{2} d \theta .\) Prove this result using the area formula with double integrals.
Gravitational field due to spherical shell A point mass \(m\) is a distance \(d\)
from the center of a thin spherical shell of mass \(M\) and radius \(R .\) The
magnitude of the gravitational force on the point mass is given by the
integral
$$F(d)=\frac{G M m}{4 \pi} \int_{0}^{2 \pi} \int_{0}^{\pi} \frac{(d-R \cos
\varphi) \sin \varphi}{\left(R^{2}+d^{2}-2 R d \cos \varphi\right)^{3 / 2}} d
\varphi d \theta$$
where \(G\) is the gravitational constant.
a. Use the change of variable \(x=\cos \varphi\) to evaluate the integral and
show that if \(d>R,\) then \(F(d)=\frac{G M m}{d^{2}},\) which means the force is
the same as if the mass of the shell were concentrated
at its center.
b. Show that if \(d
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