I am stuck in this exercise, I cannot get the right answer. The exercise is the following:
Rotate around $y = 1$ the region that is between $y=1$, $x=3$, $y=x^\frac{3}{2}$ and the x-axis.
As far as I understand, this is what I want to rotate (Ups I can't post images):
I first calculate the following:
$V_1 = \pi\int_{0}^{1}\left(x^{3/2}-1\right)^2dx = 9/20\pi$ Then I calculate the total volume of the cylinder between 0 and 1:
$V_2 = \pi \cdot1^2 \cdot1^2$
So the desired volume between 0 and 1 is:
$V = \pi - 9/20 \pi = 11/20 \pi$
Then I calculate the volume of rotating the region between 1 and 3:
$V_3 = \pi\int_{1}^{3}\left(x^{3/2}-1\right)^2dx=\pi\left(\frac{114}{5}-\frac{36\sqrt{3}}{5}\right)$
Finally the total volume:
$V_t=V_3 + V = \pi\left(\frac{467}{20}-\frac{36\sqrt{3}}{5}\right)$
However, the answer it is supposed to be:
$ANS = \pi\left(\frac{144}{5}-\frac{36\sqrt{3}}{5}\right)$
What am I doing wrong?
Firstly, I would like to echo Andre's sentiment: the way I would have interpreted the question would be to consider the area below $y=1$, above $y=0$, below $y=x^{3/2}$, and left of $x=3$. What is pictured as white in your picture. However, that result is incorrect according to your "answer".
In fact, we can see that what was intended by the answer was only the furthest right region that you plotted as red:
$$V=\pi \int_1^3 (x^{3/2}-1)^2 dx$$
which you correctly computed as "$V_3$".
However, it makes no sense to say that this region is bounded by "the x-axis," so I would approach your teacher about this problem being misleading. A better way to describe this region would be: "between $y=1$, $x=1$, $x=3$, and $y=x^{3/2}$".