Tricky volume of revolution

Question

So I was trying to find the volume, but I'm always getting wrong answer. $$x=-y^2+6y$$ and the y-axis to rotate about is $y=-7$.

$$V = \pi*\int_0^6 (-y^2+6y)^2-49 \;dy$$

But answer doesn't much? Where am I wrong?

Answer 1

The cross-section of this volume is a ring centered about $y=-7$. The ring is defined by the curves $y=f_{\pm}(x)$, where

$$f_{\pm}(x) = 3 \pm \sqrt{9-x} $$

about $y=-7$. The volume is then the sum of all the rings over $x$, which goes from $0$ to $9$:

$$V = \pi \int_0^9 dx [(7+f_+(x))^2 - (7+f_-(x))^2]$$

With a little algebra, you can show that this is

$$V = 40 \pi \int_0^9 dx \: \sqrt{9-x} = 720 \pi$$

EDIT

Here is a picture of the situation, for what it's worth:

Answer 2

You are right, it is definitely tricky. The region we are rotating has not been completely described. I take it that it is the part of the region which is simultaneously inside the parabola and inside the first quadrant. Drawing a picture is essential.

One can use slicing or cylindrical shells. We will use slicing, since that's what you attempted. Then we need to integrate with respect to $x$. Note that $x$ goes from $0$ to $9$, because the vertex of the parabola is at $(9,3)$. (It is $y=3$ because that's halfway between the roots).

We will need to solve the equation $y^2-6y+x=0$. We get $y=3\pm \sqrt{9-x}$. These are, respectively, the equations of the top half and the bottom half of the parabola.

The distance from the top point $3+\sqrt{9-x}$ to $-7$ is $10+\sqrt{9-x}$. The distance from the bottom point $3-\sqrt{9-x}$ to $-7$ is $10-\sqrt{9-x}$. So our volume is $$\pi\int_0^9 \left[(10+\sqrt{9-x})^2-(10-\sqrt{9-x})^2\right]\,dx.$$ Before integrating, square and simplify. Lots of stuff disappears, and we end up with $$\pi\int_0^9 40\sqrt{9-x}\,dx.$$ The integration of $\sqrt{9-x}$ is easy. An antiderivative is $-\frac{2}{3}(9-x)^{3/2}$. Thus the volume is $(\pi)(40)\left(\frac{2}{3}\right)(27)$.

Answer 3

Choose a rectangular horizontal strip within the region. The dimension of this chosen strip is $(-y^2+6y)dy$. The distance of this strip from the axis of rotation (which is the line $y=-7$) is given by $y+7$. Hence, the desired volume is $$V=\int_{0}^{6}2\pi(y+7)\cdot (-y^2+6y)dy=720\pi.$$