I gather you're supposed to compute the integral of $G(x,y,z)=z$ over a surface $S$ that is the part of the parabolic cylinder $y=z^2$ with $0\le x\le2$ and $0\le z\le\frac{\sqrt{15}}2$ .

We can parameterize $S$ by

$\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k$

with the given constraints on $x$ and $z$ . Take the normal vector to $S$ to be

$\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k$

so that the surface element is

$\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz$

Then in the integral, we have

$\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}$

5 0

3 years ago

PLEASE HELP asap it’s for a quiz if you know at least one answer please tell me

VARVARA [1.3K]

15
about 94.25
about 706.86

Lmk if you want to know how I got the answers

3 0

2 years ago

Challenge question<br>please help<br>

Pani-rosa [81]

Answer:

The 3 possible values of x are;

6, 4 and 2

Step-by-step explanation:

If the triangle is isosceles, then two of the sides must be equal

So we equate the sides, 2 at a time to get the different values of x

3x + 4 = 2x + 10

3x-2x = 10-4

x = 6

3x + 4 = x + 12

3x-x = 12-4

2x = 8

x = 8/2 = 4

2x + 10 = x + 12

2x - x = 12-10

x = 2

7 0

3 years ago