# Rotating curves to find volume

• Nov 25th 2009, 03:24 PM
Open that Hampster!
Rotating curves to find volume
1) Consider the solid obtained by rotating the region bounded by the given curves about the x-axis.

$y = 1-\frac{5}{2}x, y=0, x=2, x=3$

Cut it into circles and integrate the Area.

I'm assuming the y = 0 implies to ignore the bottom of the graph. From x=2 to x=3 rotating the shape around the x axi gives a sort of funnel thing with radius $A(x) = 1-\frac{5}{2}x$

Integrate it to get the volume:

$V(x) = \int^3_2 A(x)dx = \int^3_2 \pi(1-\frac{5}{2}x)^2$ = $\int^3_2 \pi(1-5x+\frac{25}{4}x^2)$

I integrated and I got $V(x) = \pi(x-\frac{5x^2}{2}+\frac{25x^3}{12})|^3_2$

Which reduced to 337/12

2) Consider the solid obtained by rotating the region bounded by the given curves about the x-axis.

$y= 16-4x^2, y=0$

Would the radius for this one just be the function and then repeat the process above?

3) Same deal as above: $y= \frac{3}{x}, x =1,x=2,y=0$

Treat the x values as boundaries for the upper half of the function and do the same thing as the first question?
• Nov 25th 2009, 03:33 PM
skeeter
Quote:

Originally Posted by Open that Hampster!
1) Consider the solid obtained by rotating the region bounded by the given curves about the x-axis.

$y = 1-\frac{5}{2}x, y=0, x=2, x=3$

Cut it into circles and integrate the Area.

I'm assuming the y = 0 implies to ignore the bottom of the graph. From x=2 to x=3 rotating the shape around the x axi gives a sort of funnel thing with radius $A(x) = 1-\frac{5}{2}x$

Integrate it to get the volume:

$V(x) = \int^3_2 A(x)dx = \int^3_2 \pi(1-\frac{5}{2}x)^2$ = $\int^3_2 \pi(1-5x+\frac{25}{4}x^2)$

I integrated and I got $V(x) = \pi(x-\frac{5x^2}{2}+\frac{25x^3}{12})|^3_2$

Which reduced to 337/12 $\textcolor{red}{\frac{337\pi}{12}}$

2) Consider the solid obtained by rotating the region bounded by the given curves about the x-axis.

$y= 16-4x^2, y=0$

Would the radius for this one just be the function and then repeat the process above? yes , r = y

3) Same deal as above: $y= \frac{3}{x}, x =1,x=2,y=0$

Treat the x values as boundaries for the upper half of the function and do the same thing as the first question? yes

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