Question

Sketch the region common to circle x^2+y^2=16 and parabola x^=6y .Also find area of region using integration

Answer 1

first solve x² + y² = 16 and x² = 6y to find intersection of both graphs.
put x² = 6y in x² + y² = 16
6y + y² = 16,
y² + 8y - 2y - 16 = 0
(y + 8)(y - 2) = 0
y = -8, 2
but y ≠ -8 because x² doesn't be negative.
so, y = 2 and x² = 6 × 2 = 12
x = ±2√3

hence, intersection points are (2√3, 2) and (-2√3, 2) as shown in figure.
Let C : x² + y² - 16 = 0 => y = √(16 - x²) [ we just took positive because enclosed area in positive y axis.]
and U : x² = 6y 0 => y = x²/6

now, from graph
area of enclosed surface = $\int\limits^{4}_{2}{\sqrt{16-x^2}}\,dx+\int\limits^2_0{\left[\frac{x^2}{6}\right]}\,dx$
solve it and get value .

Answer 2

Answer:

4/3[sqrt(3) + 4pi]

Step-by-step explanation:

In the attachments I have answered this problem.

See the attachment for detailed solution.

I hope this answer helps you