Area Integral in 2D

Calculus Level 2

Compute the area between the parabola x + 5 = y 2 x+5 = y^2 and the y y -axis.

2 3 5 1 / 2 \frac23 5^{1/2} 4 3 5 1 / 2 \frac43 5^{1/2} 2 3 5 3 / 2 \frac23 5^{3/2} 4 3 5 3 / 2 \frac43 5^{3/2}

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
Apr 24, 2016

The desired area is given by the integral:

∫ − 5 5 ∫ y 2 − 5 0 d x d y = ∫ − 5 5 ( 5 − y 2 ) d y = 10 5 − 2 3 ( 5 ) 3 = 2 ( 5 ) 3 / 2 − 2 3 ( 5 ) 3 / 2 = 4 3 ( 5 ) 3 / 2 . \int_{-\sqrt{5}}^{\sqrt{5}} \int_{y^2-5}^0 dx dy = \int_{-\sqrt{5}}^{\sqrt{5}} (5-y^2) dy = 10\sqrt{5} - \frac23 (\sqrt{5})^3 = 2(5)^{3/2} - \frac23 (5)^{3/2} = \frac43 (5)^{3/2}.

9 Helpful 1 Interesting 1 Brilliant 1 Confused

Whoops, it looks like you meant to put those things to the power of 3/2, not 2/3.

Scotty Jamison - 2 years, 11 months ago

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Fixed, thank you.

Matt DeCross - 2 years, 11 months ago

Didnt even need to integrate x.

Wallace Leung - 1 year, 3 months ago

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