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kipiarov [429]
3 years ago
9

STATE THE MODULES THAT ARE INCLUDED WHEN LEARNING CALCULUS I.

Mathematics
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

Calculus is the branch of mathematics studying the rate of change of quantities (which can be interpreted as slopes of curves) and the length, area, and volume of objects. The chain rule is a formula for the derivative of the composition of two functions in terms of their derivatives.

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Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6
Anton [14]

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

V= \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder,x^{2}+y^{2} =1 and the ellipsoid, 4x^{2}+ 4y^{2} + z^{2} =64

In polar coordinates, we know that

x^{2}+y^{2} =r^{2}

So, the ellipsoid gives

4{(x^{2}+ y^{2)} + z^{2} =64

4(r^{2}) + z^{2} = 64

z^{2} = 64- 4(r^{2})

z=± \sqrt{64-4r^{2} }

So, the volume of the solid is given by:

V= \int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta

= 2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta

To solve the integral take, 64-4r^{2} = t

dt= -8rdr

rdr = \frac{-1}{8} dt

So, the integral  \int\ r\sqrt{64-4r^{2} } rdr become

=\int\ \sqrt{t } \frac{-1}{8} dt

= \frac{-1}{12} t^{3/2}

=\frac{-1}{12} (64-4r^{2}) ^{3/2}

so on applying the limit, the volume becomes

V= 2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta

=\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta

V = \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Since, further the integral isn't having any term of \theta.

we will end here.

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

3 0
1 year ago
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