If the rectangle below has an area of 40 sq. units, what is the area of the

Express the integral as an iterated integral in six different ways, where E is the solid bounded by y=4-x^2-4z^2 and y=0

zmey [24]

Assuming you need the integral expressing the volume of $E$ , the easiest setup is to integrate with respect to $y$ first.

This is done with either

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dx\,\mathrm dz$
$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dz\,\mathrm dx$

Thanks to symmetry, integrating with respect to either $x$ or $z$ first will be nearly identical.

First, with respect to $x$ :

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dy\,\mathrm dz$
$\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dz\,\mathrm dy$

Next, with respec to $z$ :

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$
$\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dx\,\mathrm dy$

5 0

2 years ago

Find three examples of corporate logos. Do they incorporate symmetry? If so, how and what kind?

Dvinal [7]

What are you talking about?

3 0

2 years ago

Solve 9x-7y=-7 i don't know how to do it so could you show me step by step please

Lesechka [4]

First of all, you need to know what a 'solution' is, so you'll know it
when you see it.

A 'solution' to an equation is a number, or a set of numbers, that
you can write in place of the variables (the letters), and when you
do that, the equation will have only numbers in it, and it'll be a true
statement.

Your equation has two variables in it ... 'x' and 'y' . In order to find
just a single set of numbers for what both of them must be, you
would need two equations.

The way it stands now, with only one equation, there are actually
an infinite number of solutions. Each solution is a pair of numbers ...
one for 'x' and one for 'y' ... and if you write them into the equation
in place of 'x' and 'y', then the equation is a true statement.
I'll show you how to tell if a pair of numbers is a solution or not.

Here's what that looks like:

Say I give you two pairs of numbers, and I tell you that both of them
are solutions to your equation. The 'solutions' I give you are

          x=0
          y=1
and
          x=2
          y=3 .

You don't trust me, and you say to me "Wait just a minute there, dude !
Not so fast. I'll need to check them out and see if those are really solutions
to my equation."

You take the first pair and write it into your equation:
x=1, y=0
           9x - 7y = -7
                    9(0) - 7(1) = -7
                     0    - 7   = -7
                           -7   = -7
OK. That's a true statement.
So x=0, y=1 is a solution.

Now check the other one I gave you:
x=2, y=3
                   9x - 7y = -7
                    9(2) - 7(3) = -7
                       18 - 21 = -7
                          -3    = -7
This is NOT a true statement.
So x=2, y=3 is NOT a solution to your equation.
I pulled a fast one on you. If I was charging you money for solutions,
then you would not pay me for this one, because it's not a solution.

8 0

3 years ago

I need help with this please.

Setler [38]

A. 170 students out of the 230 students love both biking and skating (to make it easier, the percentage is about 0.74).
I am sorry if I cannot solve the other half of your question

5 0

2 years ago

Laura is baking a cake. The recipe says that she has to mix 64 grams of chocolate powder to the flour. Laura knows that 1 cup of

pickupchik [31]

The answer is She went over by one over six of a cup

1 cup of that particular chocolate powder has a mass of 128 grams: 1c = 128g

64 grams of chocolate powder is x cups.
1c = 128g
x = 64g

1c : 128 g = x : 64g
x = 1c : 128 g * 64 g
x = 0.5 c
64 grams of chocolate powder is 0.5 cups = 1/2 cups

She added two over three of a cup of chocolate powder: 2/3 cups

She need to add 1/2 cups: 1/2 = 3/6 cups
She added 2/3 cups: 2/3 = 4/6 cups

So she added 4/6 - 3/6 = 1/6 cups more.

5 0

3 years ago