volume of solid about y-axis

[thesheepdog]

I am new here, so please forgive any math formatting that is not up to spec.

I have the following problem that I need help with: "Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis.

y=x^2,y=0,x=0,x=3
There are 4 answers on this problem:

1) (81÷4)∗pi
2) 9pi
3) (243÷4)∗pi
4) (243÷5)∗pi
I know that to find the volume, we need to integrate these x/y values.

∫0-3 pix(x^2−0)dx
Then we take the anti-derivative:

pix4÷4
Then we calculate this over the integral by substitution:
pi(3)^4÷4−pi(0)^4÷4
So this brings us to the answer for the area in terms of pi: (81÷4)pi
Is this the correct answer?

 

[HallsofIvy]

I am new here, so please forgive any math formatting that is not up to spec.

I have the following problem that I need help with: "Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis.

y=x^2,y=0,x=0,x=3
There are 4 answers on this problem:

1) (81÷4)∗pi
2) 9pi
3) (243÷4)∗pi
4) (243÷5)∗pi
I know that to find the volume, we need to integrate these x/y values.

Unfortunately, you "know" something that isn't true. Or, more accurately, you are misremembering a formula.

If you imagine drawing a line, perpendicular to the x-axis, rotated around the x-axis, fills out a disk of radius \(\displaystyle y= x^2\) so area \(\displaystyle \pi (x^2)^2= \pi x^4\). Taking a thickness of "dx", we have a small slice of volume \(\displaystyle \pi x^4 dx\). Integrate that from 0 to 3.

∫0-3 pix(x^2−0)dx
Then we take the anti-derivative:

pix4÷4
Then we calculate this over the integral by substitution:
pi(3)^4÷4−pi(0)^4÷4
So this brings us to the answer for the area in terms of pi: (81÷4)pi
Is this the correct answer?

 

[thesheepdog]

Ah, that would make a difference.

So pi*x^4 integrated from 0 to 3 should be (243\5)*pi, correct?

 

[Deleted member 4993]

I am new here, so please forgive any math formatting that is not up to spec.

I have the following problem that I need help with: "Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis.

y=x^2,y=0,x=0,x=3
There are 4 answers on this problem:

1) (81÷4)∗pi
2) 9pi
3) (243÷4)∗pi
4) (243÷5)∗pi
I know that to find the volume, we need to integrate these x/y values.

∫0-3 pix(x^2−0)dx
Then we take the anti-derivative:

pix4÷4
Then we calculate this over the integral by substitution:
pi(3)^4÷4−pi(0)^4÷4
So this brings us to the answer for the area in terms of pi: (81÷4)pi
Is this the correct answer?

It should be:

\(\displaystyle \displaystyle{V \ = \int \pi * r^2 dx \ = \int_0^3 \pi \left [x^2\right ]^2 dx \ = \ \pi * \left |\frac{x^5}{5}\right |_0^3}\)

 

[thesheepdog]

thank you!

 

[HallsofIvy]

Ah, that would make a difference.

So pi*x^4 integrated from 0 to 3 should be (243\5)*pi, correct?

Well, I would have used "/" rather than "\"!