james scored 300 runs in cricket. he requires 900 runs to qualify for league MVP contention. he scores 50 runs during the averag
1 answer:
You might be interested in
First, complete the square in the equation for the second circle to determine its center and radius:
<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0
<em>x</em> ² - 10<em>x</em> + 25 + <em>y </em>² = 25
(<em>x</em> - 5)² + <em>y</em> ² = 5²
So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.
Now convert each equation into polar coordinates, using
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
Then
<em>x</em> ² + <em>y</em> ² = 100 → <em>r </em>² = 100 → <em>r</em> = 10
<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0 → <em>r </em>² - 10 <em>r</em> cos(<em>θ</em>) = 0 → <em>r</em> = 10 cos(<em>θ</em>)
<em>y</em> = 5 → <em>r</em> sin(<em>θ</em>) = 5 → <em>r</em> = 5 csc(<em>θ</em>)
See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to <em>y</em> = 5 at the point (0, 5).
Split up the region at 3 angles <em>θ</em>₁, <em>θ</em>₂, and <em>θ</em>₃, which denote the angles <em>θ</em> at which the curves intersect. They are
<em>θ</em>₁ = 0 … … … by solving 10 = 10 cos(<em>θ</em>)
<em>θ</em>₂ = <em>π</em>/6 … … by solving 10 = 5 csc(<em>θ</em>)
<em>θ</em>₃ = 5<em>π</em>/6 … the second solution to 10 = 5 csc(<em>θ</em>)
Then the area of the region is given by a sum of integrals:
To compute the integrals, use the following identities:
sin²(<em>θ</em>) = (1 - cos(2<em>θ</em>)) / 2
cos²(<em>θ</em>) = (1 + cos(2<em>θ</em>)) / 2
and recall that
d(cot(<em>θ</em>))/d<em>θ</em> = -csc²(<em>θ</em>)
You should end up with an area of
We can verify this geometrically:
• the area of the larger circle is 100<em>π</em>
• the area of the smaller circle is 25<em>π</em>
• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line <em>y</em> = 5, has area 100<em>π</em>/3 - 25√3
Hence the area of the region of interest is
100<em>π</em> - 25<em>π</em> - (100<em>π</em>/3 - 25√3) = 125<em>π</em>/3 + 25√3
as expected.
