Please help, I'm in trouble for not doing my homework, this is missing btw. First correct answer gets brainlist!!!!!

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

3 years ago

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2 3/4 - 1 1/8=?

Airida [17]

Basically you need to make the denominator (whats on the bottom) the same, lets take the first one for example

2 3/4, lets put this all over 4 for the same denominator.

2=8/4, and then add this to 3/4 to get 11/4.

the second part will work the same way,

start by making 1 into 8/8. add this to 1/8 to get 9/8

now we have to add 11/4 and 9/8. to do this, you have to find the smallest multiple between the two (for example: between 6 and 4 it would be 12, since 4*3= 12, and 2*6=12)

now, the smallest common multiple between 4 and 8 would just be 8

11/4= 22/8, since you multiply the bottom and top by the same amount so it still equals 11/4, but your denominator will now be 8.

now you can subtract. 22/8- 9/8= 13/8

final answer: 13/8

try the rest yourself since this is important for later math, hoped this helped :)

3 0

2 years ago

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one days discharge at its mouth, 3 trillion gallons, could supply all of country A’s household for 5 months. how much water an a

serious [3.7K]

let's use engineering notation for the sake of brevity.

1 trillion is 1,000,000,000,000, or just 1E12, twelve zeros.

1 million is then 1E6, six zeros.

we know the discharge for one day is 3E12 gallons, and that'd do just fine for country A for 5 months. How many gallons in 1 month only?

$\bf \begin{array}{ccll} gallons&months\\ \cline{1-2} 3E12&5\\ x&1 \end{array}\implies \cfrac{3E12}{x}=\cfrac{5}{1}\implies 3E12=5x \\\\\\ \cfrac{3E12}{5}=x\implies 6E11=x\implies 600000000000=x$

if there are 200million inhabitants in A, namely 200E6 or 2E8 inhabitants, how many gallons per inhabitant from all those 6E11 gallons?

$\bf \begin{array}{ccll} gallons&households\\ \cline{1-2} 6E11&2E8\\ x&1 \end{array}\implies \cfrac{6E11}{x}=\cfrac{2E8}{1}\implies 6E11=2E8x \\\\\\ \cfrac{6E11}{2E8}=x\implies \cfrac{600000000000}{200000000}=x\implies 3000=x$