Which of the following integrals correctly computes the volume formed when the region bounded by the curves x2 + y2 = 25, x = 3, and y = 0 is rotated around the y-axis?

The correct integral is    ⇒ 2nd answer

Step-by-step explanation:

The formula of the volume of a curve y = f(x) by integration around the y-axis is  

∵ The curves are x² + y² = 25 , x = 3 and y = 0

- Lets find x in terms of y in the first curve

∵ x² + y² = 25

- Subtract y² from both sides

∴ x² = 25 - y²

- Take square root for both sides

∴ x =

To find the limit of the integration b substitute x by 3

∵ x = 3

∵ x² + y² = 25

∴ 3² + y² = 25

∴ 9 + y² = 25

- Subtract 9 from both sides

∴ y² = 16

- Take square root for both sides

∴ y = ± 4

∵ The 1st limit a is at y = 0

∴ a = 0

∵ The 2nd limit is b at y = 4

∴ b = 4

∴ b = 4

There are two curves x = 3 and x =

To find the volume integrate the difference of the squares of the equations of the two curves and multiply the difference by π

∵    

- Subtract square 3 from the square of      

∴       

The correct integral is        

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!