2. Find the area of the region under the graph of f(x) = x√4-x^2
between the ordinates x=-2 and x=2. Further, use washer's
method to obtain the volume of the solid generated by revolving the
curve y = f(x) between the limits x=-2 and x = 2.
Answers
Answer:
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
Given: f(x) = √4-x^2.
we have to find out area of the curve for x = -2 to x = 2.
Given region is {(x,0):y= √4-x² } and X−axis.
We have, y= √4-x²
⇒y² = 4 + x²
⇒ x² + y² = 4.
Area of shaded region = ∫√4-x² dx = ∫√2²-x² dx
= [x/2√2²-x² + 2²/2 sin^-1(x/2)] for -2 to 2
= 2/2×0 + 2×π/2 + 2/2×0 - 2sin^-1(-1)
= 2×π/2 + 2×π/2
= π + π
= 2π sq units.
Hence, the area of the region is 2π sq units.
By washer's method
Volume, V = ∫A(x) dx
V = ∫[x/2√2²-x² + 2²/2 sin^-1(x/2)] dx for -2 to 2
Steps to find volume of the solid by washer's mehtod is
- Sketch a graph of the region to be revolved. Then revolve the region around the x-axis, sketching the result.
- Imagine slicing the solid perpendicular to the x-axis, and opening
the solid along the slice. What does the cross-section look like?
- Find an expression for the area of the cross-sectional face. For a washer, the area is given by A = πr2² - πr1²
where r2 is the larger radius and r1 is the smaller radius.
- Set up an integral representing the volume using
V =∫A(x)dx
- Evaluate the integral if required.
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