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Math Help - Rotating curves to find volume

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    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?
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  2. #2
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    Quote Originally Posted by Open that Hampster! View Post
    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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