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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

Best Answer 

 #1
avatar+120 
+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 \(\int_{0}^{8} \pi(2 - \frac{\sqrt[3]{y}}{2})^2 - \pi dy = 15.0796\)

.
 May 3, 2019
edited by Anthrax  May 3, 2019
 #1
avatar+120 
+3
Best Answer

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
avatar+322 
+1

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

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