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necessaryh

Answered question

2021-05-23

Calculate the iterated integral.

\int_{0}^{3} \int_{0}^{1} 4 x y (\sqrt{x^{2} + y^{2}}) d y d x

Answer & Explanation

Velsenw

Skilled2021-05-24Added 91 answers

Heres

Jeffrey Jordon

Expert2021-09-08Added 2605 answers

Consider the iterated integral

$\int_{0}^{3} \int_{0}^{1} 4 x y (\sqrt{x^{2} + y^{2}}) d y d x$

The objective is to calculated the iterated integral

$\int_{0}^{3} \int_{0}^{1} 4 x y (\sqrt{x^{2} + y^{2}}) d y d x$

To evaluate the iterated integral, first find the integral with respect to "y" and then apply the integral with respect to "x" $\int_{0}^{3} \int_{0}^{1} 4 x y (\sqrt{x^{2} + y^{2}}) d y d x = \int_{0}^{3} \int_{0}^{1} (2 x) (2 y) (\sqrt{x^{2} + y^{2}}) d y d x$

$= \int_{0}^{3} \int_{0}^{1} (2 x) (\sqrt{4}) d u d x$

$= \int_{0}^{3} \int_{0}^{1} (2 x) (4^{\frac{1}{2}}) d u d x$

$= \int_{0}^{3} (2 x) {(\frac{4^{\frac{1}{2} + 1}}{\frac{1}{2} + 1})}_{0}^{1} d x$

$= \int_{0}^{3} (2 x) {(\frac{4^{\frac{3}{2}}}{\frac{3}{2}})}_{0}^{1} d x$

$= \int_{0}^{3} (2 x) \cdot \frac{2}{3} (4^{\frac{3}{2}})_{0}^{1} d x$

$= \int_{0}^{3} \frac{4 x}{3} ((x^{2} + y^{2})^{\frac{3}{2}})_{0}^{1} d x$

$= \int_{0}^{3} \frac{4 x}{3} ((x^{2} + (1)^{2})^{\frac{3}{2}} - (x^{2} + (0)^{2})^{\frac{3}{2}} d x$

$\int_{0}^{3} \int_{0}^{1} 4 x y (\sqrt{x^{2} + y^{2}}) d y d x = \int_{0}^{3} \frac{4 x}{3} ((x^{2} + 1)^{\frac{3}{2}} - (x^{2})^{\frac{3}{2}}) d x$

Continuous from the last step

$\int_{0}^{3} \int_{0}^{1} 4 x y (\sqrt{x^{2} + y^{2}}) d y d x = \int_{0}^{3} \frac{4 x}{3} ((x^{2} + 1)^{\frac{3}{2}} - (x^{2})^{\frac{3}{2}}) d x$

$= \int_{0}^{3} \frac{4 x}{3} ((x^{2} + 1)^{\frac{3}{2}}) d x - \int_{0}^{3} \frac{4 x}{3} ((x^{2})^{\frac{3}{2}}) d x$

$= \int_{0}^{3} \frac{2}{3} \cdot 2 x ((x^{2} + 1)^{\frac{3}{2}}) d x - \int_{0}^{3} \frac{4 x}{3} ((x^{2})^{\frac{3}{2}}) d x$

$= \int_{0}^{3} \frac{2}{3} \cdot 2 x ((x^{2} + 1)^{\frac{3}{2}}) d x - \int_{0}^{3} \frac{4 x}{3} (x^{3}) d x$

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