Find the volume of the solid enclosed by the surfaces x² + y² + z² = a , x² + y² +z² = b , (a<b) and z = (x² + y² )½ .​

An obvious substitution would be to take u = x - y and v = x + 2y, so that 0 < u < 1 (which follows from y = x   ==>   x - y = 0 and y = x - 1   ==>   x - y = 1), and 0 < v < 1.

Solve for x and y in terms of u and v :

u - v = (x - y) - (x + 2y) = -3y   ==>   y = (v - u)/3

2u + v = 2 (x - y) + (x + 2y) = 3x   ==>   x = (2u + v)/3

Compute the Jacobian and its determinant:

Now in the integral, we have

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)