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We had double integrals over general regions and I was trying to do some practice exercises and seem to have trouble:

  1. $\int \int_A (x^3 + 1)^{\frac{1}{2}}\ dA$ with A over $0 \le x \le 2$ bounded by x-axis and $y = x^2$.

  2. $\int \int_B x^3y^2\ dA$ with B - triangle, vertices (0,0), (2,2), and (2,-2).

My attempt:

  1. $$\int_0^2 \int_0^{x^2}(x^3 + 1)^{\frac{1}{2}} \ dydx. $$ For the inside integral I got $(x^3 + 1)^{\frac{1}{2}}y \ |_0^{x^2} = (x^3 + 1)^{\frac{1}{2}} x^2$ with the outside one I'm having a problem.

2.$$\int_0^2 \int_{-x}^x x^3y^2 \ dydx$$ For the inside integral I got $ x^3 \frac{y^3}{3} |_{-x}^x= \frac{2x^6}{3}$ and for the outside and the final answer $ \frac{2}{3} \frac{x^7}{7}|_0^2 =\frac{256}{21}$

Would appreciate any help.

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    $\begingroup$ For the first problem, for the outside integral use the substitution $u=x^3+1$. $\endgroup$ – André Nicolas Mar 31 '14 at 3:38
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    $\begingroup$ I believe you have correctly computed the second integral. $\endgroup$ – Guest Mar 31 '14 at 3:40
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In the first problem, the setup as an iterated integral is correct. For the outside integral, use the substitution $u=x^3+1$. Then $du=3x^2\,dx$, and we want $$\int_{u=1}^9 \frac{1}{3}u^{1/2}\,du.$$ The integration is straightforward.

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  • $\begingroup$ I was just calculating that: and thankfully, got the same #s :) $\frac{1}{3} \frac{2}{3}u^{3/2} \ |_1^9 = \frac{52}{9}$ $\endgroup$ – Bob Mar 31 '14 at 3:51
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    $\begingroup$ Yes, that is right. $\endgroup$ – André Nicolas Mar 31 '14 at 3:52

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