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Chapter 13: Q.29 (page 991)

If the density at each point in $T$ is proportional to the point's distance from the $x$ -axis, find the center of mass of $T$ .

Short Answer

Expert verified

Area of the region bounded by the spiral and the $x$ -axis is

$A = \frac{π^{3}}{6}$

Step by step solution

01

given information

The objective of this problem is to find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral.

The region is enclosed by the spiral $r = θ$ and the $x$ -axis on the interval localid="1650634059460" $0 \leq θ \leq π$

Plot of $r = θ$

Area of the region bounded by the spiral and the $x$ - axis can be expressed as

$A = \int_{0}^{x} \int_{0}^{r - θ} r d r d θ$

Integrate with respect to $r$ first.

$A = \int_{0}^{*} {[\frac{r^{2}}{2}]}_{0}^{e} d θ A = \int_{0}^{π} [\frac{θ^{2} - 0}{2}] d θ A = {[\frac{θ^{3}}{6}]}_{0}^{*}$

Area of the region bounded by the spiral and the $x$ -axis is

$A = \frac{π^{3}}{6}$

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Most popular questions from this chapter

Let $Ω$ be a lamina in the xy-plane. Suppose $Ω$ is composed of n non-overlapping laminæ role="math" localid="1650321722341" $Ω_{1}, Ω_{2}, \dots, Ω_{n} .$ Show that if the masses of these laminæ are $m_{1}, m_{2}, \dots, m_{n}$ and the centers of masses are $({\bar{x}}_{1}, {\bar{y}}_{1}), ({\bar{x}}_{2}, {\bar{y}}_{2}), \dots, ({\bar{x}}_{n}, {\bar{y}}_{n}),$ then the center of mass of $Ω$ is $(\bar{x}, \bar{y}),$ where

$\bar{x} = \frac{\sum_{k = 1}^{n} m_{k} {\bar{x}}_{k}}{\sum_{k = 1}^{n} m_{k}} and \bar{y} = \frac{\sum_{k = 1}^{n} m_{k} {\bar{y}}_{k}}{\sum_{k = 1}^{n} m_{k}} .$

Evaluate each of the double integral in the exercise 37-54 as iterated integrals

$\int \int_{R} \sin (x + 2 y) d A W h e r e R = {(x, y) | 0 \leq x \leq π, 0 \leq y \leq \frac{π}{2}}$

Find the signed volume between the graph of the given function and the xy-plane over the specified rectangle in the xy-plane

$f (x, y) = x y W h e r e R = {(x, y) | - 2 \leq x \leq 3, - 1 \leq y \leq 5}$

Let $Ω$ be a lamina in the xy-plane. Suppose $Ω$ is composed of two non-overlapping lamin $Ω_{1}$ and $Ω_{2}$ , as follows:

Show that if the masses and centers of masses of $Ω_{1}$ and $Ω_{2}$ are $m_{1}$ and $m_{2},$ and $({\bar{x}}_{1}, {\bar{y}}_{1}) and ({\bar{x}}_{2}, {\bar{y}}_{2})$ respectively, then the center of mass of $Ω$ is $(\bar{x}, \bar{y}),$ where

$\bar{x} = \frac{m_{1} {\bar{x}}_{1} + m_{2} {\bar{x}}_{2}}{m_{1} + m_{2}} and \bar{y} = \frac{m_{1} {\bar{y}}_{1} + m_{2} {\bar{y}}_{2}}{m_{1} + m_{2}}$

Let f(x, y, z) and g(x, y, z) be integrable functions on the rectangular solid $R = \{(x, y, z) | a_{1} ⩽ x ⩽ a_{2}, b_{1} ⩽ y ⩽ b_{2}, c_{1} ⩽ z ⩽ c_{2}\}$ . . Use the definition of the triple integral to prove that :

$∭_{R} (f (x, y, z) + g (x, y, z)) d V = ∭_{R} f (x, y, z) d V + ∭_{R} g (x, y, z) d V .$

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