Mathematics

Mathematics

Problem

Let TT be the region bounded by y=x2y=x^{2}, the xx-axis, and the line x=2x=2. When TT is revolved about the horizontal line y=−1y=-1, what is the exact volume of the solid generated? a) 206π15\dfrac{206\pi}{15} b) 176π15\dfrac{176\pi}{15} c) 136π15\dfrac{136\pi}{15} d) 96π15\dfrac{96\pi}{15}
Intuitive explanation

When you revolve a region around a horizontal line that's not the xx-axis, the key is to shift your perspective so the axis of rotation becomes your reference point. Here, revolving around y=−1y = -1 means every point in TT is actually a distance of (its yy-coordinate plus 1) away from the axis. Use the washer method with outer radius R(x)=x2−(−1)=x2+1R(x) = x^2 - (-1) = x^2 + 1 and inner radius r(x)=0−(−1)=1r(x) = 0 - (-1) = 1, then integrate the area of each washer from x=0x = 0 to x=2x = 2.

Intuitive explanation

When you revolve a region around a horizontal line that's not the xx-axis, the key is to shift your perspective so the axis of rotation becomes your reference point. Here, revolving around y=−1y = -1 means every point in TT is actually a distance of (its yy-coordinate plus 1) away from the axis. Use the washer method with outer radius R(x)=x2−(−1)=x2+1R(x) = x^2 - (-1) = x^2 + 1 and inner radius r(x)=0−(−1)=1r(x) = 0 - (-1) = 1, then integrate the area of each washer from x=0x = 0 to x=2x = 2.

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