The area formed in the first quadrant by the graphs of 𝑦=𝑥2 and 𝑦=8−𝑥2 is revolved about the y -axis. Using a vertical element, dV is equal to
Question
The area formed in the first quadrant by the graphs of 𝑦=𝑥2 and 𝑦=8−𝑥2 is revolved about the y -axis. Using a vertical element, dV is equal to
Solution
The volume of a solid of revolution can be found using the formula:
dV = π(R^2 - r^2)dy
where R is the outer radius and r is the inner radius.
In this case, the outer radius R is the distance from the y-axis to the curve y = 8 - x^2 and the inner radius r is the distance from the y-axis to the curve y = x^2.
Since we are revolving around the y-axis, we need to express x in terms of y for each function.
For y = x^2, we have x = sqrt(y) and for y = 8 - x^2, we have x = sqrt(8 - y).
So, the outer radius R = sqrt(8 - y) and the inner radius r = sqrt(y).
Substituting these into the formula gives:
dV = π[(sqrt(8 - y))^2 - (sqrt(y))^2]dy = π[(8 - y) - y]dy = π[8 - 2y]dy
So, the differential volume element dV is equal to π[8 - 2y]dy.
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