# Volume of the solid plane

#### ozgunozgur

##### New member
Hello there, can you help me to solve this problem? Thanks.

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#### Prof

##### Junior Member
You need to submit an attempt at solving the problem, and we can help you where you are stuck. In the mean time, I can get you started.
In both parts, start by making a graph of the 3 equations and shade the region that is bounded by them.
Then both questions can be solved using the disk method for volume.

#### ozgunozgur

##### New member
You need to submit an attempt at solving the problem, and we can help you where you are stuck. In the mean time, I can get you started.
In both parts, start by making a graph of the 3 equations and shade the region that is bounded by them.
Then both questions can be solved using the disk method for volume.
Should I rotate it separately for three of them?

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#### MarkFL

##### Super Moderator
Staff member
Let's look at a graph of the region to be revolved:

For part (a), I would use the washer method, where the outer radius $$R$$ is the parabolic function and the inner radius $$r$$ is the linear function. Can you give the volume of an arbitrary washer?

$$\displaystyle dV=\pi(R^2-r^2)=?$$

#### ozgunozgur

##### New member
I don't know if this is true. Thanks

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#### MarkFL

##### Super Moderator
Staff member
For part (a) we are revolving about the $$x$$-axis. I would write:

$$\displaystyle dV=\pi((x^2+4)^2-(4x)^2)\,dx=\pi(x^4-8x^2+16)\,dx$$

Now integrate over the appropriate boundaries. What do you get?

I did

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#### MarkFL

##### Super Moderator
Staff member
You found the anti-defivative of the integrand, but you need to evaluate a definite integral.

#### ozgunozgur

##### New member
You found the anti-defivative of the integrand, but you need to evaluate a definite integral.
Sorry but I didn't understand which boundaries I need to take. Could you solve all?

#### MarkFL

##### Super Moderator
Staff member
Look at the diagram I posted above. We are essentially revolving vertical strips about the $$x$$-axis, where these strips are bounded by the two given functions (quadratic is top boundary, and linear is bottom boundary). The width of each strip is $$dx$$. What is the smallest and largest value for $$x$$ over which we are adding washers (integrating)? Can you express the volume as a definite integral and then use the FTOC to compute the value?

You need to be able to do this yourself, as I am happy to guide you, but you'll learn very little if I just provide the answers.

#### ozgunozgur

##### New member
I wonder. Must the smallest and largest value be equal to the zero and eight like this or the strips are already those functions? However, I like to learn with examination solvings. And I don't want to keep you hours here.

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#### MarkFL

##### Super Moderator
Staff member
No, the strips run from $$x=0$$ to $$x=2$$. We are given the lower bound, since the vertical line $$x=0$$ is one of the boundaries of the shaded region. The upper limit fomrs from there the quadratic and linear functions intersect:

$$\displaystyle x^2+4=4x$$

$$\displaystyle x^2-4x+4=0$$

$$\displaystyle (x-2)^2=0\implies x=2$$

Hence:

$$\displaystyle V=\pi\int_0^2x^4-8x^2+16\,dx=\frac{256}{15}\pi$$

#### ozgunozgur

##### New member
No, the strips run from $$x=0$$ to $$x=2$$. We are given the lower bound, since the vertical line $$x=0$$ is one of the boundaries of the shaded region. The upper limit fomrs from there the quadratic and linear functions intersect:

$$\displaystyle x^2+4=4x$$

$$\displaystyle x^2-4x+4=0$$

$$\displaystyle (x-2)^2=0\implies x=2$$

Hence:

$$\displaystyle V=\pi\int_0^2x^4-8x^2+16\,dx=\frac{256}{15}\pi$$
Thank you so much, sir. At the other part(y-axis), will the boundaries need again 0 and 2, and x will be equal to 2?

#### MarkFL

##### Super Moderator
Staff member
When revolving about the $$y$$-axis, I would use the shell method, where:

$$\displaystyle dV=2\pi x((x^2+4)-4x)\,dx$$

Do you see that the radius of an arbitrary shell is $$x$$ and the height is the difference between the quadratic and the line?

#### ozgunozgur

##### New member
When revolving about the $$y$$-axis, I would use the shell method, where:

$$\displaystyle dV=2\pi x((x^2+4)-4x)\,dx$$

Do you see that the radius of an arbitrary shell is $$x$$ and the height is the difference between the quadratic and the line?
I found 8 pi. Is this correct? What is the difference between washer and shell method?

#### Subhotosh Khan

##### Super Moderator
Staff member
I found 8 pi. Is this correct? What is the difference between washer and shell method?

Volume by washer method

Volume by shell method

Tell us what you found and tell us if you do not understand something in any of those references.

#### MarkFL

##### Super Moderator
Staff member
I found 8 pi. Is this correct? What is the difference between washer and shell method?
Let me check:

$$\displaystyle V=2\pi\int_0^2 x((x^2+4)-4x)\,dx=2\pi\int_0^2 x^3-4x^2+4x\,dx=2\pi\left[\frac{1}{4}x^4-\frac{4}{3}x^3+2x^2\right]_0^2=2\pi\left(4-\frac{32}{3}+8\right)=\frac{8}{3}\pi$$