What are the methods for finding volume using washers and shells?

[Justabeginner]

Homework Statement

The region in the first quadrant bounded by the lines y=4 and x=2 and the curve x^2=4y is revolved about the y-axis. Find the volume:
a) by the method of washers
b) by the method of shells

Homework Equations

The Attempt at a Solution

I drew the region, and I can picture what the solid volume would look like. I think the limits of the integral are 0 and 4, if I'm not mistaken. Also, for the washer method, I need the inner and outer radius. I think the outer radius is x^2/4, but I'm not sure what the inner radius would be. As for the shell method, I think the radius would be 4 and the height, I'm not so sure about. Any help is appreciated. Thank you so much.

 

[QED Andrew]

Here is a graph of the region in question. Note the region in question is the region above the parabola (and below y = 4), not the region below the parabola (and above y = 0).

You correctly identified the limits of integration for the washer method; however, the upper and lower limits of integration for the shell method are x = 0 and x = 2, respectively. I suggest you review the formulas associated with the washer and shell methods.

For the washer method, the radius $R(y)$ is given by

$$R(y) = \left \{ \begin{array}{lr} 2\sqrt{y} , & 0 < y < 1\\ 2 , & 1 < y < 4 \end{array} \right.$$

Because the radius is a piece-wise function, two integrals must be evaluated to find the volume.

For the shell method, the radius $p(x)$ is given by $p(x) = x$ and the height $h(x)$ is given by $h(x) = 4 - \dfrac{x^2}{4}$.

After you review the washer and shell methods, you should be able to progress from here.

 

[Justabeginner]

I do not understand how you determined there to be two integrals for the radius for the washer method. If you could please explain that to me, I'd appreciate it. Thank you.

 

[QED Andrew]

By the washer method, the formula for the volume $V$ is given by

$$V = \pi \int_0^4 [R(y)]^2 \, dy$$

Keeping in mind $R(y)$ is a piecewise-defined function with two subfunctions, can you evaluate $\int_0^4 [R(y)]^2 \, dy$ with only one integral?

 

[Justabeginner]

No you would need two integrals, one with a limit from 0 to 1, and then the other integral with a limit from 1 to 4.

For the first integral:

$∫_0^1 4y dy$
$2y^2 | _0^1$
$2∏$

Second Integral:

$∫_0^1 4 dy$
$4y | _1^4$
12∏

12∏ + 2∏= 14∏

The same answer as I got for the shell method. I understand how you got the 2√y for the function of the first integral, but I do not get how you got the '2' for the second integral. Can you please explain that to me? Thank you so much.

 

[QED Andrew]

I attached a graphic to aid our discussion. The black line at $y = 1$ appropriately divides the graph into two regions, one below the line and one above the line. In each region, an arbitrary radius is represented by a red rectangle. In the lower region where $0 < y < 1$, the radius $R(y)$ is given by $R(y) = 2\sqrt{y}$ as you already know. In the upper region where $1 < y < 4$, what is the radius $R(y)$? From the attached graphic, it should be clear $R(y)$ is constant in this region.

 

[Justabeginner]

Thank you so much. The graph really helped me understand it better.