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AURORKA [14]
2 years ago
7

Why do I measure capacity

Mathematics
1 answer:
earnstyle [38]2 years ago
7 0
To find the maximum for something pertaining to a liquid or etc.
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\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:

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\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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