Comparing two lines - Ancova: lm or aov?

Hello!

I have measurements (length and volume) of fish collected in two years. I
want to know if the the relationship between length and volume is the same
for both years. The number of fish measured is different for each year. I
don't know whether lm or aov is more appropriate to use.

Here are the two output options:

Call:
lm(formula = Volume ~ Length * Year)

Residuals:
? ? Min ? ? ? 1Q ? Median ? ? ? 3Q ? ? ?Max
-1.35062 -0.42695 -0.02234 ?0.29789 ?1.78540

Coefficients:
? ? ? ? ? ? ? ? ?Estimate Std. Error t value Pr(>|t|)
(Intercept) ? ? ?-8.699125 ? 0.543372 -16.010 ? <2e-16 ***
Length ? ? ? ? ? ?0.151462 ? 0.006708 ?22.578 ? <2e-16 ***
Year1 ? ? ? ? 1.627903 ? 1.168158 ? 1.394 ? 0.1660
Length:Year1 -0.024259 ? 0.014510 ?-1.672 ? 0.0972 .
---
Signif. codes: ?0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.6016 on 119 degrees of freedom
Multiple R-squared: 0.8384, ? ? Adjusted R-squared: 0.8343
F-statistic: 205.8 on 3 and 119 DF, ?p-value: < 2.2e-16

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AND for aov:
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? ? ? ? ? ? ? Df ?Sum Sq Mean Sq ?F value ? ?Pr(>F)
Length ? ? ? ? ?1 219.457 219.457 606.4349 < 2.2e-16 ***
Year ? ? ? ? ?1 ? 2.970 ? 2.970 ? 8.2073 ?0.004934 **
Length:Year ? 1 ? 1.012 ? 1.012 ? 2.7952 ?0.097173 .
Residuals ? ? 119 ?43.064 ? 0.362
---
Signif. codes: ?0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1


Thanks very much for any help.
Kirsten

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