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LUCKY_DIMON [66]

3 years ago

13

The length of a running trail in the park is 12 miles.There are water fountains every 1/4 mile.How many fountains are along the

running trail?

1 answer:

Alex73 [517]3 years ago

8 0

There are 4 foutains along the runnong trail

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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:

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3 0

1 year ago

Write an inequality for <br> A farmer weighs a dozen chicken eggs. The heaviest egg is 56g.

almond37 [142]

Answer:

672

Step-by-step explanation: all you have to do is multiplication

3 0

2 years ago

Read 2 more answers

Factoring without Combining Like Terms

Alex_Xolod [135]

Answer:

(x + 3)(2x + 5)

Step-by-step explanation:

Given

2x² + 6x + 5x + 15 ← grouping the terms

= (2x² + 6x) + (5x + 15) ← factor each group

= 2x(x + 3) + 5(x + 3) ← factor out (x + 3) from each term

= (x + 3)(2x + 5) ← in factored form

8 0

2 years ago

If a function has a vertical asymptote at a certain x-value, then the function is _____ at that value.

Dmitriy789 [7]

The denominator of the function would be equal to zero so it would be undefined.

8 0

2 years ago

Read 2 more answers

Given T(x,y) = ( x + h, y + k) and the point P(5, 2), what is T (P) = (x + 7, y - 3)? Answer should be a coordinate point.

timofeeve [1]

Answer:

<h3>(12, -1)</h3>

Step-by-step explanation:

Given T(x,y) = ( x + h, y + k) and the point P(5, 2), we are to find T (P) = (x + 7, y - 3)

It can be seen from the given data that T (P) = T (x, y), hence P = (x, y)

Given the coordinate P(5, 2), comparing with P(x, y), x =5 and y =2

Substituting x = 5 and y = 2 into the coordinates T (P) = (x + 7, y - 3)

T (P) = (5+7, 2-3)

T (P) = (12, -1)

Hence the coordinate T(P) = (12, -1)

5 0

2 years ago