2

The functions are $x=\frac{y^2}{2}$, $x=2$, and $y=2$. I graphed it and it looks like the intersection points are $(2,2)$ and $(0,2)$.

But I don't know how to set up the integral.

DiegoMath
  • 4,059
Pia
  • 95
  • So now that you have the graph, slice it up along the $y$-axis (easy since you have the graph given as $x=g(y)$), revolve that slice around the $x$-axis, and find the volume of the shell. Then add up all the shells (ie turn it into an integral). – tabstop Feb 15 '14 at 18:22
  • I have my integral set up (2y - (y^3)/(2y). This is after distributing the y. But I'm not sure how to find the antiderivative for this. – Pia Feb 15 '14 at 18:40
  • I don't see where the $y$ came from in the denominator; but even with it, it just cancels out of the top. – tabstop Feb 15 '14 at 19:35
  • The y came from the shell method formula. But yes I see that they would cancel out! However, I plug two into the integral of y^3 and get 4. And 4 times 2pi is 8pi. The answer is 4pi. So I'm still not sure what I'm doing wrong. – Pia Feb 15 '14 at 19:41
  • You don't have $y^3$, you have $1-y^2/2$. (Assuming I'm reading your notation correctly.) – tabstop Feb 15 '14 at 19:43

1 Answers1

1

For the shell method, finding the volume when rotating about the $x$-axis, we have the general formula:

$$\int_a^b 2\pi\,r_y\,h_y\,dy$$

In this case, we have $a = 0, b= 2$, $r_y = y$ ("radius"), and height $h_y$ given by the curve $h_y = \frac {y^2}{2}$.

That gives us $$\int_0^2 2\pi\, y\,\left(\frac {y^2}{2}\right)\,dy \quad= \quad\pi\int_0^2 y^3\,dy\quad = \quad \frac{\pi}4 y^4\,\Big]_0^2$$

amWhy
  • 209,954