# Thread: Rotating curves to find volume

1. ## 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?

2. 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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