1
Question
Draw a rough sketch of the region {(x,y) : y2≤6ax & x2+y2≤16a2}. Also find the area of the region sketched using method of integration.
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Solution
We have y2≤6ax
⇒y2=6ax
is a corresponding parabola ... (1)
x2+y2≤16a2
⇒x2+y2=16a2
is a corresponding circle ... (2)
On solving (1) and (2), we get
⇒x2+6ax−16a2=0
⇒(x−2a)(x+8a)=0
x=2a (Accepted) or x=−8a (Rejected as its outside circle)
So, the area bounded by the circle and parabola is shaded in the diagram below:
Required area is twice the sum of area bounded by y2=6ax and x axis when x varies from 0 to 2a and area bounded by x2+y2=16a2 and x−axis, when x varies from 2a to 4a, because the bounded region is symmetric about x−axis as both circle and parabola are symmetric about x−axis.
Area =∫x2x1(y2−y1)dx
⇒ Area =2⎡⎢⎣∫2a0(√6ax−0)dx+∫4a2a(√(4a)2−x2−0)dx⎤⎥⎦
[∴x varies from 0 to 4a]
⇒ Area =2⎡⎢⎣√6a×23x32⎤⎥⎦2a0
+2⎡⎢
⎢⎣lx2√(4a)2−x2+16a22sin−1x4a⎤⎥
⎥⎦4a2a
⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣∴∫√a2−x2dx=x2√a2−x2+a22sin−1xa+c∫baxndx=[xn+1n+1]ba⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
=2⎡⎢
⎢
⎢
⎢⎣l√6a×23(2a)32−0+8a2×π2−2a2√16a2−4a2−8a2×π6⎤⎥
⎥
⎥
⎥⎦
[∴sin−11=π2,sin−112=π6]
=2[8a2√33+4a2π−2√3a2−4a2π3]
=2[23×√3a2+8a2π3]
=43a2(√3+4π) sq.units
Hence, the required area is
43a2(√3+4π) sq.umits
⇒y2=6ax
is a corresponding parabola ... (1)
x2+y2≤16a2
⇒x2+y2=16a2
is a corresponding circle ... (2)
On solving (1) and (2), we get
⇒x2+6ax−16a2=0
⇒(x−2a)(x+8a)=0
x=2a (Accepted) or x=−8a (Rejected as its outside circle)
So, the area bounded by the circle and parabola is shaded in the diagram below:
Required area is twice the sum of area bounded by y2=6ax and x axis when x varies from 0 to 2a and area bounded by x2+y2=16a2 and x−axis, when x varies from 2a to 4a, because the bounded region is symmetric about x−axis as both circle and parabola are symmetric about x−axis.
Area =∫x2x1(y2−y1)dx
⇒ Area =2⎡⎢⎣∫2a0(√6ax−0)dx+∫4a2a(√(4a)2−x2−0)dx⎤⎥⎦
[∴x varies from 0 to 4a]
⇒ Area =2⎡⎢⎣√6a×23x32⎤⎥⎦2a0
+2⎡⎢ ⎢⎣lx2√(4a)2−x2+16a22sin−1x4a⎤⎥ ⎥⎦4a2a
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣∴∫√a2−x2dx=x2√a2−x2+a22sin−1xa+c∫baxndx=[xn+1n+1]ba⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
=2⎡⎢ ⎢ ⎢ ⎢⎣l√6a×23(2a)32−0+8a2×π2−2a2√16a2−4a2−8a2×π6⎤⎥ ⎥ ⎥ ⎥⎦
[∴sin−11=π2,sin−112=π6]
=2[8a2√33+4a2π−2√3a2−4a2π3]
=2[23×√3a2+8a2π3]
=43a2(√3+4π) sq.units
Hence, the required area is
43a2(√3+4π) sq.umits
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