volume by evaluating π ( y^2 )^2 dy = π (5y^4 ) dy. We get π (5y^4 ) dy = π(y^5 ) = 2π.- 8 π
When the region we are revolving is defined between a curve f(x) and g(x), we can find thevolume using cylindrical shells. We use the formula V = 2π x[f(x)− g(x)] dx. Here we havef(x) = x and g(x) = − . Thus, the height of each shell is f(x)− g(x) = x − , andthe radius is simply x. The endpoints of our region are x = 0 and x = 2. Therefore, we will findthe volume by evaluating 2π dx = 2π dx = 3π x^2 dx. We get 3π x^2dx = 3π = 8π.6.
When the region we are revolving is defined between a curve f(x) and g(x), we can find thevolume using cylindrical shells. We use the formula V = 2π x[f(x)− g(x)] dx. Here we havef(x) = and g(x) = 2x − 1. Thus, the height of each shell is f(x)− g(x) = −(2x − 1) = − 2x + 1, and the radius is simply x. The left endpoint of our region is x = 0, and we find theright endpoint by finding where f(x) = intersects g(x) = 2x − 1. We get x = 1. Therefore,we will find the volume by evaluating = . We get =
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7.
When the region we are revolving is defined between a curve f(y) and g(y), we can find the