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2 years ago

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Mathematics

1 answer:

Vilka [71]2 years ago

4 0

$83.60 in gross wages

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Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y =

Strike441 [17]

Answer:

The volume is:

$\displaystyle\frac{37\pi}{10}$

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

$\displaystyle\int_a^b\pi r^2dx$

Notice here the radius of the washer is the difference of the given curves:

$x-\sqrt{x}$

So the integral becomes:

$\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx$

We solve it:

Factor $\pi$ out and distribute the exponent (you can use FOIL):

$\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx$

Notice: $x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$

So the integral becomes:

$\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx$

Then using the basic rule to evaluate the integral:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4$

Simplifying a bit:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4$

Then plugging the limits of the integral:

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Taking the root (rational exponents):

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Then doing those arithmetic computations we get:

$\displaystyle\frac{37\pi}{10}$

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3 years ago

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Alina [70]

18 x 3 + 12 - 1 x 34 + 13 - 2 x 56 = 67

<=Work=>

18 x 3 = 54

1 x 34 = 34

2 x 56 = 112

(54) + 12 - (34) + 13 - (112)

66 - 34 + 13 - 112

32 + 13 - 112

45 - 112

= 67

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