1
$\begingroup$

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):

enter image description here

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?

$\endgroup$
1
  • 1
    $\begingroup$ My interpretation of the problem is that the white first quadrant region below the red is being rotated. But that does not give the official answer. $\endgroup$ Commented Aug 26, 2015 at 16:30

1 Answer 1

0
$\begingroup$

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 $y=x^{3/2}$".

$\endgroup$
1
  • $\begingroup$ Yes, I agree with both of you, I tried to do the other approach since that y = 0 is misleading. However I never got the "right answer", I think the teacher is wrong. Going tho accept this answer. $\endgroup$
    – dpalma
    Commented Aug 26, 2015 at 22:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .