# Finding the Volume of a Solid.

#### shamieh

##### Active member
Find the Volume of a Solid by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

$$\displaystyle x = y^2$$, $$\displaystyle x = 1 - y^2$$, about $$\displaystyle x = 3$$

So here is how far I've gotten with this problem. I need help though. Any guidance will be greatly appreciated.. One of the problems is I don't understand what "about x = 3" means exactly? Like what do they mean "about x = 3?" Anyways here is what I have:

I have a right and left parabola basically. I set the two equal to each other to find lines of intersection.

$$\displaystyle y^2 = 1 - y^2$$

thus;
$$\displaystyle y = \frac{\sqrt{2}}{{2}}$$

So I know the areas are even symmetry so I can say $$\displaystyle 2\pi \int^\frac{\sqrt{2}}{2}_0 (Right - left)dy$$

right?

so am i correct in saying $$\displaystyle 2\pi \int^\frac{\sqrt{2}}{2}_0 (1 - y^2)^2 - (y^2)^2 dy$$?

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#### MarkFL

Staff member
Here is a plot:

Imagine the plot rotated 90° clockwise. Now the outer radius of an arbitrary washer will be the distance between the axis of rotation $x=3$ and the function $x=y^2$, while the inner radius will be the distance from the axis of rotation and the function $x=1-y^2$.

You are correct that you can use symmetry as follows:

$$\displaystyle V=2\pi\int_0^{\frac{1}{\sqrt{2}}} (\text{outer radius})^2-(\text{inner radius})^2\,dy$$

#### shamieh

##### Active member
$$\displaystyle 2\pi \int ^\sqrt{2/2}_0 (3 - y^2)^2 - (3 - (1 - y^2)^2 dy$$

$$\displaystyle 2\pi \int^\sqrt{2/2}_0 9 - 6y^2 + y^4 - 3 + 1 + 2y^2 - y^4$$

$$\displaystyle = 2\pi \int^\sqrt{2/2}_0 5 - 8y^2 + 2y^4$$

Is this correct?

#### MarkFL

Staff member
$$\displaystyle 2\pi \int ^\sqrt{2/2}_0 (3 - y^2)^2 - (3 - (1 - y^2)^2 dy$$

$$\displaystyle 2\pi \int^\sqrt{2/2}_0 9 - 6y^2 + y^4 - 3 + 1 + 2y^2 - y^4$$

$$\displaystyle = 2\pi \int^\sqrt{2/2}_0 5 - 8y^2 + 2y^4$$

Is this correct?
Your upper limit should be $$\displaystyle \frac{1}{\sqrt{2}}$$ or $$\displaystyle \frac{\sqrt{2}}{2}$$.

Now, your integrand in the first line is correct, except for a missing closing bracket. After that you have made some algebra mistakes. You should initially have:

$$\displaystyle \left(3-y^2 \right)^2-\left(3-\left(1-y^2 \right) \right)^2$$

$$\displaystyle \left(3-y^2 \right)^2-\left(3-1+y^2 \right)^2$$

Collect like terms:

$$\displaystyle \left(3-y^2 \right)^2-\left(2+y^2 \right)^2$$

Expand:

$$\displaystyle \left(9-6y^2+y^4 \right)-\left(4+4y^2+y^4 \right)$$

Distribute negative sign and remove brackets:

$$\displaystyle 9-6y^2+y^4-4-4y^2-y^4$$

Collect like terms:

$$\displaystyle 5-10y^2=5\left(1-2y^2 \right)$$

And so you may state:

$$\displaystyle V=10\pi\int_0^{\frac{1}{\sqrt{2}}} 1-2y^2\,dy$$

#### shamieh

##### Active member
Thanks for helping me with my signs Mark. Seems like that was the hardest part of the whole problem lol.

so

$$\displaystyle 10 * (y - \frac{2y^3}{3}) | 0 to \frac{\sqrt{2}}{2}$$

$$\displaystyle \frac{30\pi\sqrt{2}}{6} - \frac{10\pi}{6} = \frac{10\pi\sqrt{2}}{3}$$

WOW! What a problem. Is this correct? I'm praying it is

#### MarkFL

$$\displaystyle V=10\pi\left[y-\frac{2}{3}y^3 \right]_0^{\frac{1}{\sqrt{2}}}= \frac{10\pi}{\sqrt{2}}\left(1-\frac{2}{3}\cdot\frac{1}{2} \right)= \frac{10\pi}{\sqrt{2}}\cdot\frac{2}{3}= \frac{20\pi}{3\sqrt{2}}= \frac{10\sqrt{2}\pi}{3}$$