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Nookie1986 [14]
4 years ago
13

Three cylinders have a height of 8 cm. Cylinder 1 has a radius of 1 cm. Cylinder 2 has a radius of 2 cm. Cylinder 3 has a radius

of 3 cm. Find the volume of each cylinder.
Mathematics
2 answers:
Bingel [31]4 years ago
6 0

Answer:

Step-by-step explanation:

The volume formula for a cylinder is B*h

B : the area of the base

h : the height

Cylinder 1

The volume is 8pi

Cylinder 2

The volume is 32pi

Cylinder 3

The volume is 72pi

QveST [7]4 years ago
4 0

Answer:

Cylinder 1:  V = 8π

Cylinder 2:  V = 32π

Cylinder 3:  V = 72π

Step-by-step explanation:

The volume of the cylinder of radius r and height h is calculated using the form:  V = π r² h

Cylinder 1:  r = 1 cm  and  h = 8 cm

Volume of Cylinder 1 = π r² h = π 1² * 8 = 8π

Cylinder 2:  r = 2 cm  and  h = 8 cm

Volume of Cylinder 2 = π r² h = π 2² * 8 =  π * 4 * 8 = 32π

Cylinder 3:  r = 3 cm  and  h = 8 cm

Volume of Cylinder 1 = π r² h = π 3² * 8 = π * 9 * 8 = 72π

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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
2 years ago
Name the prism in each case given that all lateral faces are rectangles and the bases are: Equilateral triangles
Cloud [144]
This is a triangular prism.

The lateral faces are triangles; this means that the two bases are parallel to one another.  This makes it a prism instead of a pyramid.

The bases being triangles makes it a triangular prism.
6 0
3 years ago
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