Curvy Calculus

Calculus Level 4

For F ⃗ = y x 3 i ^ + y 2 j ^ \vec{F}=yx^{3}\hat{i}+y^{2}\hat{j}

Find : ∫ C F ⃗ . d r ⃗ \int_{C}\vec{F}.d\vec{r} on the portion of the curve , x 2 = y x^{2}=y from ( 0 , 0 ) (0,0) to ( 1 , 1 ) (1,1)


The answer is 0.5.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Otto Bretscher
Feb 15, 2016

Since y = x 2 y=x^2 , we have ∫ C y x 3 d x + y 2 d y = ∫ 0 1 ( x 2 x 3 + x 4 2 x ) d x = ∫ 0 1 3 x 5 d x = 3 6 [ x 6 ] 0 1 = 0.5 \int_{C}yx^3dx+y^2dy=\int_{0}^{1}(x^2x^3+x^42x)dx=\int_{0}^{1}3x^5dx=\frac{3}{6}[x^6]_{0}^{1}=\boxed{0.5}

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