Answer:
ello mate my day is great how about yours
Step-by-step explanation:
and may i ask are you british??
The equation in spherical coordinates will be a constant, as we are describing a spherical shell.
r(φ, θ) = 8 units.
<h3>
How to rewrite the equation in spherical coordinates?</h3>
The equation:
x^2 + y^2 + z^2 = R^2
Defines a sphere of radius R.
Then the equation:
x^2 + y^2 + z^2 = 64
Defines a sphere of radius √64 = 8.
Then we will have that the radius is a constant for any given angle, then we can write r, the radius, as a constant function of θ and φ, the equation will be:
r(φ, θ) = 8 units.
If you want to learn more about spheres, you can read:
brainly.com/question/10171109
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:
and now the derivative of this expression with respect to "t" is:
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
is continuous on [a,b], differentiable on (a,b) and 
Since this this function 
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
Answer:
3y= -2x-6
Step-by-step explanation:
The two points on the line are (-3,0) and (0,-2)
so you first get the gradient;
gradient= <u>change in y</u>
change in x
= <u>-2-0</u>
0-(-3)
=<u> -2</u>
3
so the answer above is the gradient
then pick one point that you used to get the gradient with, so as for me I'll pick (-3,0) and then a general point which is always (x, y)
since you have the gradient you can easily get the equation by doing this
<u>y-0</u><u>. </u><u> </u> = <u>-2</u>
x-(-3). 3
then crossmultiply to get the equation of the line
3y= -2(x+3)
3y= -2x-6
