Chapter 6: Q. 32 (page 556)

The region between $f (x) = x^{2} \sqrt{3 - x}$ and the x-axis on [a, b] = [0, 3], with n = 6.

Short Answer

Expert verified

$C e n t r e o f m a s s : (1, 2)$

Step by step solution

Step1. Given Information

region,

$f (x) = x^{2} \sqrt{3 - x}$

Step2. Calculate

C.O.M

$f (x) = 6 - 2 x [0, 3] 0 = 6 - 2 x 2 x = 6 x = \frac{6}{2} x = 3$

Step3. Continue

$M_{x} = \frac{1}{2} P \int_{0}^{3} {(6 - 2 x)}^{2} d x M_{x} = P \int_{0}^{3} 18 - 6 x + 2 x^{2} d x M_{x} = P {[18 x - \frac{6 x^{2}}{2} + \frac{2 x^{3}}{3}]}_{0}^{3} M_{x} = P [54 - 54 + \frac{54}{3}] = 18 P$

Step4. Continue

$M_{y} = P \int_{0}^{3} x (6 - 2 x) d x M_{y} = P \int_{0}^{3} 6 x - 2 x^{2} d x M_{y} = P {[\frac{6 x^{2}}{2} - \frac{2 x^{3}}{3}]}_{0}^{3} M_{y} = P [27 - \frac{54}{3}] M_{y} = P [27 - 18] = 9 P$

Step4. Continue

$M = P \int_{0}^{3} (6 - 2 x) d x = 6 x - x^{2} {|_{0}}^{3} = P [18 - 9] = 9 P$

Step6. Continue

$M = 9 P, M_{y} = 9 P, M_{x} = 18 P C . o . M = (X_{c m}, Y_{c m}) = (\frac{M_{y}}{M}, \frac{M_{X}}{M}) = (\frac{9 P}{9 P}, \frac{18 P}{9 P}) = (1, 2)$

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The region between $f (x) = x^{2} \sqrt{3 - x}$ and the x-axis on [a, b] = [0, 3], with n = 6.

Short Answer

Step by step solution

Step1. Given Information

Step2. Calculate

Step3. Continue

Step4. Continue

Step4. Continue

Step6. Continue

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The region between f(x)=x23-xand the x-axis on [a, b] = [0, 3], with n = 6.

Short Answer

Step by step solution

Step1. Given Information

Step2. Calculate

Step3. Continue

Step4. Continue

Step4. Continue

Step6. Continue

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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The region between $f (x) = x^{2} \sqrt{3 - x}$ and the x-axis on [a, b] = [0, 3], with n = 6.