Thread: Need help finding the volume(shells or washers method)

1. Need help finding the volume(shells or washers method)

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid using any method.

y = x^2 + x - 2, y = 0, about the x-axis.

I'm lost on how to do this. The graph of x^2 + x - 2 would go under the graph and when you intersect that with y = 0, shouldn't you get a negative answer when you integrate it from root -1.61803 to 0.618034? But the book says the answer is 81pi/10.

2. 0 - ( a negative) = a positive.

AND/OR

(negative)^2 = positive

3. First of all, I don't understand why you're integrating from $\displaystyle \displaystyle -1.61803$ to $\displaystyle \displaystyle 0.618034$.

The function is a quadratic, with zeroes at $\displaystyle \displaystyle x=-2$ and $\displaystyle \displaystyle x = 1$. So the region that is being rotated is the region between these values.

It doesn't matter that this function's values are negative in this region, because it is their absolute values that's important.

The volume of this solid of revolution can be approximated by summing cylinders of radius $\displaystyle \displaystyle |y|$, so the circular cross-sectional area is $\displaystyle \displaystyle \pi |y|^2 = \pi y^2$, and height $\displaystyle \displaystyle dx$, where $\displaystyle \displaystyle dx$ is some small change in $\displaystyle \displaystyle x$.

So $\displaystyle \displaystyle V \approx \sum{\pi y^2\,dx} = \sum{\pi (x^2 + x - 2)^2\,dx}$.

As you increase the number of cylinders and $\displaystyle \displaystyle dx \to 0$, this sum converges on an integral, and this approximation becomes exact, so the integral you are evaluating is

$\displaystyle \displaystyle \int_{-2}^1{\pi (x^2 + x - 2)^2 \,dx}$

$\displaystyle \displaystyle = \pi\int_{-2}^1{x^4 + 2x^3 - 3x^2 - 4x + 4\,dx}$

$\displaystyle \displaystyle = \pi \left[\frac{x^5}{5} + \frac{x^4}{2} - x^3 - 2x^2 + 4x\right]_{-2}^1$

$\displaystyle \displaystyle = \pi \left[\left(\frac{1}{5} + \frac{1}{2} - 1 - 2 + 4\right) - \left(-\frac{32}{5} + 8 + 8 - 8 - 8\right)\right]$

$\displaystyle \displaystyle = \frac{81\pi}{10}$.

4. Thank you Prove It. I just noticed I wrote down the question wrong on my paper. I wrote down y = x^2 + x - 1 instead of y = x^2 + x - 2. No wonder I couldn't get the answer.

Anyways could you help me on another problem.

x^2 + (y-1)^2 = 1; about the y-axis

If I solve for y, I get y = sqrt(1-x^2) + 1

And if you graph that equation you would see that the points does not even touch the x-axis. How do you find the limits for this problem so that you can integrate it. The answer should be 4pi/3.

5. You don't even need calculus for this problem, you should realise that this is the graph of a circle, and when you rotate a circle, you get a sphere.

What is the volume of a sphere?

6. Originally Posted by florx
Thank you Prove It. I just noticed I wrote down the question wrong on my paper. I wrote down y = x^2 + x - 1 instead of y = x^2 + x - 2. No wonder I couldn't get the answer.

Anyways could you help me on another problem.

x^2 + (y-1)^2 = 1; about the y-axis

If I solve for y, I get y = sqrt(1-x^2) + 1

And if you graph that equation you would see that the points does not even touch the x-axis. How do you find the limits for this problem so that you can integrate it. The answer should be 4pi/3.
Let me ask you this: can we not rotate half the circle (i.e. a semi-circle) about the y-axis and multiply by two? And if we can, does the volume of a semi-circle being rotated about the y-axis differ from one that similarly lyes on the x-axis and is also rotated about the x-axis?

7. Yes we can. But I must show work on how to do this problem with integrals. My teacher checks the steps so I can't just say that and be over with it. The instructions is same as the problem above, "The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid using any method."

8. OK, if you need to do this as a solid of revolution, notice that this time you need have cylinders stacked vertically, not horizontally as in the previous problem. This means we need to know the $\displaystyle \displaystyle y$ bounds. Note that $\displaystyle \displaystyle 0 \leq y \leq 2$, so we will be integrating over that region.

Also note that if $\displaystyle \displaystyle x^2 + (y-1)^2 = 1$

$\displaystyle \displaystyle x^2 = 1 - (y - 1)^2$

$\displaystyle \displaystyle x^2 = 1 - (y^2 - 2y + 1)$

$\displaystyle \displaystyle x^2 = 2y - y^2$.

The radius of each cylinder is $\displaystyle \displaystyle x$, which means the area of each circular cross-section is $\displaystyle \displaystyle \pi x^2$, and the height of each cylinder is $\displaystyle \displaystyle dy$, a small change in $\displaystyle \displaystyle y$.

So the volume can be approximated by $\displaystyle \displaystyle V \approx \sum{\pi x^2 \,dy} = \sum{\pi (2y - y^2)\,dy}$.

As you increase the number of cylinders and $\displaystyle \displaystyle dy \to 0$, the sum converges on an integral and the approximation becomes exact.

So $\displaystyle \displaystyle V = \pi \int_0^2{2y - y^2\,dy}$.

9. Thank you so much for your time and great explanation on how to solve this problem!