I'm studying for a test and am stuck on one volume problem.
The figure is rotated about the x-axis and is bounded by $$ y=x^2, y=\frac{(3-x)}{2},y=0. $$
I thought the solution would be: $$ \pi\int_{0}^{1}(\frac{3-x}{2})^{2}-(x^{2})^{2}dx $$ However, the key is indicating that this is wrong. Any insights as to where I may have gone wrong would be appreciated.
Additional question: Is the area integrated between the two curves and the x-axis then rotated about the x-axis? Or is the area integrated between the two curves and the y axis then rotated about the x-axis? Which area am I "shading?" Unsure if that made sense but I believe that is where I'm fundamentally misunderstanding the problem
hints
$f(x)= x^2.$
$g(x) = (3-x)/2.$
where does:
$f(x)$ hit the $x$-axis?
$g(x)$ hit the $x$-axis?
$f(x)$ hit $g(x)$?
given a height $f(x)$, from $a$ to $b$, what is the formula for the volume of revolution generated by $f(x)$?
Addendum
Per OP's comment/question:
The stated problem actually involves a region that has one height for $x \leq 1$ and a different height for $x \geq 1.$ This is how you should be interpreting the instructions:
The figure is rotated about the x-axis and is bounded by $$ y=x^2, y=\frac{(3-x)}{2},y=0. $$
This means that the volume of revolution is completely solid, with no (inner gap). This means that the cross-sectional area is $\pi\times R^2$, where $R$ is either equal to $f(x)$ or $g(x)$, depending on whether $x \leq 1$ or $x \geq 1$.
Contrast that with a different problem:
Let $s(x) = 2.$
Let $t(x) = 1.$
Consider the region bounded by $x=0$, $x=3$, $s(x)$ and $t(x)$.
If you were considering the volume of revolution for this region rotated about the $x$-axis, you would realize that the 3 dimensional solid formed has a hole in it, like a hollow cylinder.
Here, the formula for the cross-sectional area would be $\pi \times (R^2 - r^2)$, where $R$ is the outer radius of $s(x)$ and $r$ is the inner radius of $t(x)$.