I am in Calculus and I need to find the volume of a solid created by an area revolved around the y axis.
So the problem I have is $y=x^2$, $y=2-x$, $x=0$, for $x>=0$ and I need to find the volume of the solid that is created by rotating around the Y axis.
The shell method is
$V= \int_a^b 2\pi(shell radius)(shell height)dx$.
It seems when I graph this, the area that is bound by $y=x^2$, $y=2-x$ and the x axis – creating a small triangle on the bottom – is what is going to form the solid. I guess my first question is whether or not this is a correct assumption?
If that is the case, I think I need to create two integrals. The integral of 0 to 1 and another of 1 to 2.
The integrals would be:
$V= \int_1^0 2\pi(x)(x^2)dx$
$V= \int_1^2 2\pi(x)(2-x)dx$
The reason I think I need two integrals is because the height of the shape is based on $y=x^2$ when x is between 0 and 1 then the height is based on $y=2-x$ when x is between 1 and 2.
Assuming this is true, these volumes would need to be added together because each is only half of the problem.
The first volume is:
$V= 2\pi \int_0^1 x^3 dx$
The antiderivative of $x^3$ is $x^4/4$
$V = 2\pi ((1^4/4)-(0^4/4))$
$V = 2\pi ((1/4) – (0/4))$
$V = 2\pi/4$
$V = \pi/2$
The second volume is:
$V= 2\pi \int_1^2 2x-x^2 dx$
The antiderivative of $2x-x^2$ is $x^2-x^3/3$
$V = 2\pi((2^2-2^3/3)-(1^2-1^3/3))$
$V = 2\pi((4-8/3)-(1-1/3))$
$V = 2\pi((4/3)-(2/3))$
$V = 2\pi(2/3)$
$V = 4\pi/3$
$V = \pi/2 + 4\pi/3$
$V = 3\pi/6 + 8\pi/6 = 11\pi/6$
$V = 5.76$ (rounded to the nearest hundredth)
I have been at this problem too long and I have myself twisted so if this is right great, if I am wrong can you provide some assistance where I went wrong?
Best Answer
Yes, you have correctly identified the area to rotate, and yes, it does require two separate integrals, since the "height" function (upper function - lower function) is different on the intervals $[0, 1]$ and $[1, 2]$. You appear to have set up the integrals and calculated them correctly.