Evaluate \int_{0}^{1/2}\int_{\sqrt{3}y}^{\sqrt{1 - y^2}}xy^2\ dxdy.

Question:

Evaluate {eq}\int_{0}^{1/2}\int_{\sqrt{3}y}^{\sqrt{1 - y^2}}xy^2\ dxdy. {/eq}

Double Integration:

In this case, we have to evaluate the double integrals over the given intervals. The general form of the double integration method is {eq}\displaystyle \text{I}=\int_{y_{1}}^{y_{2}}\int_{x_{1}}^{x_{2}} f(x, y) \ dx dy {/eq}.

The given function is integrating twice with respect to the variables for the intervals.

Answer and Explanation:

Evaluating the Integral:

{eq}\begin{align*} \\ \displaystyle \text{I} &=\int_{0}^{\frac{1}{2}}\int_{\sqrt{3}y}^{\sqrt{1-y^2}}(xy^2) \ dxdy \\ \\ \displaystyle &=\int_{0}^{\frac{1}{2}} \left[ y^2 \cdot \frac{x^{1+1}}{1+1} \right]_{\sqrt{3}y}^{\sqrt{1-y^2}} \ dy \\ \\ \displaystyle &=\int_{0}^{\frac{1}{2}} \left[ y^2 \cdot \frac{x^{2}}{2} \right]_{\sqrt{3}y}^{\sqrt{1-y^2}} \ dy \\ \\ \displaystyle &=\int_{0}^{\frac{1}{2}} \left[ \left( y^2 \cdot \frac{\left( \sqrt{1-y^2} \right)^{2}}{2} \right)-\left( y^2 \cdot \frac{\left( \sqrt{3}y \right)^{2}}{2} \right) \right] \ dy \\ \\ \displaystyle &=\int_{0}^{\frac{1}{2}} \left( \frac{y^2-4y^4}{2} \right) \ dy \\ \\ \displaystyle &=\frac{1}{2} \int_{0}^{\frac{1}{2}} \left( y^2-4y^4 \right) \ dy \\ \\ \displaystyle &=\frac{1}{2} \left[ \frac{y^{2+1}}{2+1}-4\cdot \frac{y^{4+1}}{4+1} \right]_{0}^{\frac{1}{2}} \\ \\ \displaystyle &=\frac{1}{2} \left[ \frac{y^{3}}{3}-4\cdot \frac{y^{5}}{5} \right]_{0}^{\frac{1}{2}} \\ \\ \displaystyle &=\frac{1}{2} \left[ \left( \frac{\left( \frac{1}{2} \right)^{3}}{3}-4\cdot \frac{\left( \frac{1}{2} \right)^{5}}{5} \right)-\left( \frac{(0)^{3}}{3}-4\cdot \frac{(0)^{5}}{5} \right) \right] \\ \\ \displaystyle &=\frac{1}{2} \left[ \frac{1}{60} \right] \\ \\ \displaystyle \text{I} &=\frac{1}{120} \end{align*} {/eq}

Therefore, the integrated answer is {eq}\ \displaystyle \mathbf{\color{blue}{ I =\frac{1}{120} }} {/eq}.


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Double Integration: Method, Formulas & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 15
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