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# Volume of Functions

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y= 8x^3 , y=0 , x=1 , about x=2

Not sure how but I got the volume to equal 0 so I clearly did something wrong.

May 3, 2019

#1
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For this question, integrating with respect to y is the way to go >:]

Defining R and r

If we have a general volume with the vertical axis of revolution being x = c with the formula $$\int_{a}^{b} \pi*R^2 - \pi*r^2 dy$$, we can define $$R = (c - g(y))^2$$ and $$r = (c - f(y))^2$$

However, in this case, r stays a constant value of 1 distance away from x = 2, so $$r = 1$$

Now we should define our outer function, $$y = 8x^3$$, in terms of y to be g(y).

$$y = 8x^3 \rightarrow \frac{y}{8}=x^3 \rightarrow \sqrt[3]{\frac{y}{8}}=x$$ or more simply $$g(y) = \frac{\sqrt[3]{y}}{2}$$

Therefore, $$R = \pi*(2 - \frac{\sqrt[3]{y}}{2})^2$$

Finding Bounds of Integration

Our original x defined bounds of integration were [0,1], so let's redefine with our new g(y) equation set to 0 and 1.

$$0 = \frac{\sqrt[3]{y}}{2} \rightarrow y = 0$$ and $$1 = \frac{\sqrt[3]{y}}{2} \rightarrow 2 = \sqrt[3]{y} \rightarrow y=8$$

So, our new bounds of integration are [0,8].

Computing the Integral

Our integral then becomes $$\int_{0}^{8} \pi(2 - \frac{\sqrt[3]{y}}{2})^2 - \pi dy$$. I'm a little too lazy to work out the integral in LaTeX right now lol so $$\int_{0}^{8} \pi(2 - \frac{\sqrt[3]{y}}{2})^2 - \pi dy = 15.0796$$

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May 3, 2019
edited by Anthrax  May 3, 2019

#1
+115
+3

For this question, integrating with respect to y is the way to go >:]

Defining R and r

If we have a general volume with the vertical axis of revolution being x = c with the formula $$\int_{a}^{b} \pi*R^2 - \pi*r^2 dy$$, we can define $$R = (c - g(y))^2$$ and $$r = (c - f(y))^2$$

However, in this case, r stays a constant value of 1 distance away from x = 2, so $$r = 1$$

Now we should define our outer function, $$y = 8x^3$$, in terms of y to be g(y).

$$y = 8x^3 \rightarrow \frac{y}{8}=x^3 \rightarrow \sqrt[3]{\frac{y}{8}}=x$$ or more simply $$g(y) = \frac{\sqrt[3]{y}}{2}$$

Therefore, $$R = \pi*(2 - \frac{\sqrt[3]{y}}{2})^2$$

Finding Bounds of Integration

Our original x defined bounds of integration were [0,1], so let's redefine with our new g(y) equation set to 0 and 1.

$$0 = \frac{\sqrt[3]{y}}{2} \rightarrow y = 0$$ and $$1 = \frac{\sqrt[3]{y}}{2} \rightarrow 2 = \sqrt[3]{y} \rightarrow y=8$$

So, our new bounds of integration are [0,8].

Computing the Integral

Our integral then becomes $$\int_{0}^{8} \pi(2 - \frac{\sqrt[3]{y}}{2})^2 - \pi dy$$. I'm a little too lazy to work out the integral in LaTeX right now lol so $$\int_{0}^{8} \pi(2 - \frac{\sqrt[3]{y}}{2})^2 - \pi dy = 15.0796$$

Anthrax May 3, 2019
edited by Anthrax  May 3, 2019
#2
+322
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Thank you.  Whenever the object is rotating around an axis that's not the x-axis or y-axis I always get confused.

Ruublrr  May 3, 2019