Q.Find the area of the region bounded by the circle x2+y2=16 and the line x=2, lying to the right of the line x=2.
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✓ Free question
Concept understanding — Area between Two Curves
When two curves y=f(x) and y=g(x) cross each other, they enclose a bounded region between them. To find its area, the first step is always to locate the points where the two curves meet, by solving their equations simultaneously -- these intersection points become the limits of integration, and they are almost never given directly in the problem statement.
Once the limits x=a and x=b are known, the enclosed area is the difference between the area under the upper curve and the area under the lower curve, over that same interval, taken as a positive quantity:
A=∫abf(x)dx−∫abg(x)dx
In practice this means identifying which of the two curves lies above the other throughout the interval (usually by testing a convenient point between the intersection points), subtracting the lower function from the upper one, and integrating that difference directly -- rather than integrating each curve separately against the X-axis and subtracting afterwards, which is equivalent but usually more work.
This idea covers a wide range of problems that look different on the surface but share the same method: the region between two parabolas, the region between a parabola and a straight line, the segment of a circle or an ellipse cut off by a chord, and the sector of a circle cut off by a line through the centre (where the region is naturally split into a triangular piece and a curved piece, each found by a different technique, and then added together). Sketching both curves and marking their intersection points before integrating is the single most reliable way to avoid setting up the wrong difference of integrals.
Use the circular-segment formula (or integrate directly) for the region right of x=2 inside x2+y2=16.
✓Final answer
The area is 316π−43 square units.
This is the minor segment of the circle cut off by the chord x=2, at perpendicular distance d=2 from the centre, radius r=4.
The circle x2+y2=16 has radius r=4. The region to the right of x=2 (both above and below the x-axis) is bounded by the arc and the chord x=2:
Area=2∫2416−x2dx=2[2x16−x2+8sin−1(4x)]24.
At x=4: 24(0)+8sin−1(1)=0+8⋅2π=4π.
At x=2: 2216−4+8sin−1(21)=12+8⋅6π=23+34π.
∫2416−x2dx=4π−(23+34π)=312π−4π−23=38π−23.
Doubling for both the upper and lower halves of the region:
Area=2(38π−23)=316π−43.
(As a check, this matches the circular-segment formula r2cos−1(d/r)−dr2−d2 with r=4,d=2: 16cos−1(21)−212=16⋅3π−43=316π−43.)
✓Final answer
The area is 316π−43 square units.
Integrate 2∫drr2−x2dx using the standard antiderivative, recognising sin−1(1/2)=π/6 as a standard angle; cross-check against the circular-segment formula.
Using the wrong standard angle for sin−1(1/2) (it is π/6, not π/3 or π/4); forgetting to double the single-branch integral for the region above and below the x-axis.