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Area between x=g(y) and y Axis

The area boun...

Question

The area bounded by $x = 0, x = 6 + 5 y - y^{2}$ is

\frac{517}{6}

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\frac{278}{3}

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\frac{280}{3}

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\frac{343}{6}

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Solution

The correct option is D $\frac{343}{6}$

$x = 0 \Rightarrow 6 + 5 y - y^{2} = 0 \Rightarrow y^{2} - 5 y - 6 = 0$
$\Rightarrow (y - 6) (y + 1) = 0$
$\Rightarrow y = - 1 o r 6$
Required area = $\int_{- 1}^{6} (6 + 5 y - y^{2}) d y = {[6 y + \frac{5 y^{2}}{2} - \frac{y^{3}}{3}]}_{1}^{6}$
$= [36 + 90 - 72] [- 6 + \frac{5}{2} - \frac{1}{3}] = \frac{343}{6}$

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Similar questions

y = x^{2}, x = y^{2}

A

A

x = 4 - y^{2}

Y

B

A

x = 6 + 5 y - y^{2}

Y

C

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2 x = y^{2} - 1

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o f A, B, C

x^{2} + y^{2} \leq 2 a x, y^{2} \geq a x

x \geq 0

y \leq 0

x = y^{2}

y = x - 6

x^{2} + y^{2} \leq 2 a x

y^{2} \geq a x

, a > 0, x > 0, Y > 0

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