Washer/cylindrical shell volume help!

[OhhCalculus]

So basically I don't (really) have a clue what I'm doing. I get the gist of this, but not entirely. Any input would be great.

I need to find the volume "in the first quadrant."

y=x^2, y=9, about the y-axis

A. 9/2?
B. 81/2
C. 81/2?
D. 81/4?
E. 9?

Those are the possible answers. Any work, however, or some explaining on what I should do first even, would help a lot. Thanks!

 

[galactus]

You can go about this with shells or washers.

Washers may be easiest in this case.

Using washers, we will integrate w.r.t y from 0 to 9.

Since \(\displaystyle y=x^{2}\Rightarrow x=\pm\sqrt{y}\)

Since we are in the first quadrant, we use \(\displaystyle x=\sqrt{y}\)

\(\displaystyle \pi\int_{0}^{9}(\underbrace{\sqrt{y}}_{\text{r}})^{2}\overbrace{dy}^{\text{h}}\)

With washers, see what the formula resembles?. The area of a cylinder formula, \(\displaystyle {\pi}r^{2}h\).

What is a washer, but a very flat cylinder with a hole in the center. An annulus.

Each slice is the volume of an 'infinitely thin' cylinder, then we add them all up.

Using shells, the cross sections are parallel to the axis we are revolving about.

So, we integrate w.r.t x from 0 to 3.

Shells gives the volume of two concentric right circular cylinders with inner and outer radius, and height h.

\(\displaystyle 2\pi\cdot [\text{average radius}]\cdot [\text{height}]\cdot [\text{thickness}]\)

The shell has thickness dx, height f(x), and average radius x.

\(\displaystyle 2\pi\int_{0}^{3}x\cdot (\underbrace{9-x^{2}}_{\text{height}})\overbrace{dx}^{\text{thickness}}\)

Here is a graph to help see what it looks like.

 

[OhhCalculus]

Wow, that was much more help than I was expecting. Thank you so much.

And I ended up with 81/2?, or C, does that seem correct?

 

[galactus]

Yep. \(\displaystyle \frac{81\pi}{2}\)