NI - 16 2 Which of the following is equivalent to O 6 08 O 12 O 64

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

3 years ago

The length of a rectangle is 1 units less than the width. The area of the rectangle is 20 units. What is the width, in units, of

Degger [83]

Answer:

5

Step-by-step explanation:

20 has many factors, including 4&5.

4 is one less than 5, so 4 is the length and 5 is our width.

5 0

3 years ago

I'm not sure how to start the process....