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# College calculus problem very difficult help.?

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line

x=-3

y=x^2, x=y^2

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- The Integral ∫Lv 71 decade agoFavorite Answer
y = x^2

y = +/- √(x)

A (x) = π ( outer radius )^2 - π ( inner radius )^2

A (x) = π ( 3 + √(x) )^2 - π ( 3 + x^2 )^2

A (x) = π ( 9 + 6√(x) + x ) - π ( 9 + 6x^2 + x^4 )

A (x) = π [ 9 + 6√(x) + x - 9 - 6x^2 - x^4 ]

A (x) = π [ 6√(x) + x - 6x^2 - x^4 ]

1

∫ π [ 6√(x) + x - 6x^2 - x^4 ] dx

0

1

π [ 6 * (2/3) * x^(3/2) + (1/2)x^2 - 2x^3 - (1/5)x^5 ]

0

π [ 4 ( 1^(3/2) - 0 ] + (1/2)[ 1^2 - 0 ] - 2 [ 1^3 - 0 ] - (1/5)[ 1^5 - 0 ]

π [ 4 + (1/2) - 2 - (1/5) ] = π23/10

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