A spherical solid, centered at the origin, has radius 4 and mass density(x,y,z)=6-(x^2+y^2+z^2). Set up the triple integral and find its mass.
I've attached a photo of the question.
Answers
There's something very off about this question.
In spherical coordinates,
x² + y² + z² = ρ²
so that
f(x, y, z) = 6 - (x² + y² + z²)
transforms to
g(ρ, θ, φ) = 6 - ρ²
When transforming to spherical coordinates, we also introduce the Jacobian determinant, so that
dV = dx dy dz = ρ² sin(φ) dρ dθ dφ
Since we integrate over a sphere with radius 4, the domain of integration is the set
E = {(ρ, θ, φ) : 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π}
so that the integral is
[tex]\displaystyle \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 - \rho^2) \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]
Computing the integral is simple enough.
[tex]\displaystyle = \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]
[tex]\displaystyle = 2\pi \int_{\phi=0}^{\phi=\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\phi[/tex]
[tex]\displaystyle = 2\pi \left(\int_{\phi=0}^{\phi=\pi} \sin(\phi) \, d\phi\right) \left(\int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \, d\rho\right)[/tex]
[tex]\displaystyle = 2\pi \cdot 2 \cdot \left(-\frac{384}5\right) = \boxed{-\frac{1536\pi}5}[/tex]
but the mass can't be negative...
Chances are good that this question was recycled without carefully changing all the parameters. Going through the same steps as above, the mass of a spherical body with radius R and mass density given by
[tex]\delta(x, y, z) = k - (x^2 + y^2 + z^2)[/tex]
for some positive number k is
[tex]\dfrac{4\pi r^3}{15} \left(5k - 3r^2\right)[/tex]
so in order for the mass to be positive, we must have
5k - 3r² ≥ 0 ⇒ k ≥ 3r²/5
In this case, k = 6 and r = 4, but 3•4²/5 = 9.6.
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Answer:
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I HOPE THIS IS HELPING
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See below (along with unit circle attached)
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It really helps to use the unit circle! I've attached it below for your convenience. Keep in mind that a point on the unit circle is defined by [tex](cos(\theta),sin(\theta))[/tex] where [tex]\theta[/tex] is measured in radians.
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[tex]\displaystyle{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]
Answer :[tex]\displaystyle{\boxed{\red{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}\:=\:\dfrac{127\:\sqrt{3}}{6}}}}[/tex]
Step-by-step-explanation:
We have given an expression.
We have to simplify the expression.
The given expression is
[tex]\displaystyle{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]
[tex]\displaystyle{\implies\sf\:4\:\sqrt{16\:\times\:3}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]
We know that,[tex]\displaystyle{\boxed{\pink{\sf\:\sqrt{a\:\times\:b}\:=\:\sqrt{a}\:\times\:\sqrt{b}\:}}\:\cdots\sf\:a\:,\:b\:\geq\:0}[/tex]
[tex]\displaystyle{\implies\sf\:4\:\sqrt{16}\:\times\:\sqrt{3}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]
[tex]\displaystyle{\implies\sf\:4\:\times\:4\:\sqrt{3}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}}[/tex]
We know that,[tex]\displaystyle{\boxed{\blue{\sf\:\sqrt{\dfrac{a}{b}}\:=\:\dfrac{\sqrt{a}}{\sqrt{b}}\:}}\:\cdots\sf\:b\: > \:0}[/tex]
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[tex]\displaystyle{\implies\sf\:\dfrac{8\:\times\:4\:\times\:3\:-\:5\:+\:(\:6\:\sqrt{3}\:\times\:2\:\sqrt{3}\:)}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:-\:5\:+\:(\:6\:\times\:2\:\times\:\sqrt{3}\:\times\:\sqrt{3}\:)}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:-\:5\:+\:(\:12\:\times\:3\:)}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:-\:5\:+\:12\:\times\:3}{2\:\sqrt{3}}} \\ \\ \\ \displaystyle{\implies\sf\:\dfrac{8\:\times\:12\:+\:12\:\times\:3\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{12\:(\:8\:+\:3\:)\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{12\:\times\:11\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{132\:-\:5}{2\:\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{127}{2\:\sqrt{3}}} \\ \\ \\ \displaystyle{\implies\sf\:\dfrac{127}{2\:\sqrt{3}}\:\times\:\dfrac{\sqrt{3}}{\sqrt{3}}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{127\:\sqrt{3}}{2\:\times\:3}} \\ \\ \\\displaystyle{\implies\sf\:\dfrac{127\:\sqrt{3}}{6}} \\ \\ \\ \displaystyle{\therefore\:\underline{\boxed{\red{\sf\:4\:\sqrt{48}\:-\:\dfrac{5}{2}\:\sqrt{\dfrac{1}{3}}\:+\:6\:\sqrt{3}\:=\:\dfrac{127\:\sqrt{3}}{6}}}}}[/tex]
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Answer:
f-1(x) = 1/3x would be the answer
Step-by-step explanation:
Answer:
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Answer:
D) [tex]\left[\begin{array}{cc}-1&2&1&9\end{array}\right][/tex]
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What is the answer to y=-2(-3/2)-9
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Answer:
-6
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Hey there!
y= -2 (-3/2) - 9
-2(-3/2) = [3]
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y = -6
Therefore, your answer is: y = -6
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)
PLEASE HELP WILL MARK BRAINLIEST!
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Step-by-step explanation:
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Answer:
rational number
Step-by-step explanation:
Natural numbers:
This is every number in the sequence 1, 2, 3, 4...
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This is all of the above, but it also includes 0.
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All of the above, but this also includes negative numbers.
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Basically anything that wasn't covered above. These cannot be represented as a fraction of 2 integers.
[tex]\pi, \sqrt{3}[/tex]
=========================
You don't make it clear if this is a repeating decimal or not, but either way, this is a rational number as you can represent it as a fraction:
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In the case that it was a repeating decimal, those are also rational numbers. Take this for example:
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Answer:
log14=1.146
log18=1.255
Step-by-step explanation:
Use log rules to break up the given logs, to be able to use the information give
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1. x^2 +2x -24 = 0
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Step-by-step explanation:
to round 4.645 to the nearest whole number consider the tenths’ value of 4.645, which is 6 and equal or more than 5. Therefore, we have to round up: the whole number part of 4.645, 4, increases by 1 to 5, and the decimal point and all digits (.645) are removed.
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Answer:
75%, or 7
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Step-by-step explanation:
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x=4
Step-by-step explanation:
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a
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Answers
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I NEED HELP!!
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Answers
Answer:
f(x)=x(3/2)
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6
Find the least common denominator for these fractions. Enter your answer in
the space provided.
Answers
Answer: 27
Step-by-step explanation: LCM of 3 and 27 is 27.