Question

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{1} \int_{y}^{\sqrt{y}} x^{2} y^{2} d x d y $$

Short answer

The value of the double integral is \(\frac{1}{42}\)

Step-by-step solution

Understand the given integral and the region of integration

The given double integral \(\int_{0}^{1}\int_{y}^{\sqrt{y}} x^{2}y^{2}dxdy\) represents area calculation on a region of xy-plane which has bounds 0 to 1 for y-coordinate and y to\(\sqrt{y}\) for x-coordinate.

Visualization of the area of integration

The limits of integration set the region in the xy-plane: \(0 \leq y \leq 1\) and \(y \leq x \leq \sqrt{y}\). This region represents a part of the xy-plane where y values range from 0 to 1, and for each fixed y, x ranges from y to \(\sqrt{y}\). This is a somewhat warped triangle-shaped region in the first quadrant.

Change The Order of Integration

Changing the order of integration is often a good strategy when the given limits are 'awkward' or complicated. The given limits are somewhat difficult to deal with, so consider switching the order. If the range for x is redefined as \(0 \leq x \leq 1\), the range for y can be defined as \(x^{2} \leq y \leq x\). This simplifies things since now, x is ranging from 0 to 1 and y is ranging within values of x.

Compute the double integral

Evaluating the new integral: \(\int_{0}^{1}\int_{x^{2}}^{x} x^{2}y^{2}dydx\). This is split into two separate integrals. Firstly, compute the inner integral with respect to y: \(\int_{x^{2}}^{x} x^{2}y^{2}dy = [x^{2} \cdot \frac{y^{3}}{3}]_{x^{2}}^{x}\), which simplifies to \(\frac{1}{3}(x^{5}-x^{6})\). Then the outer integral is \(\int_{0}^{1} \frac{1}{3}(x^{5}-x^{6})dx = [\frac{1}{6}\cdot x^{6} - \frac{1}{7}\cdot x^{7}]_{0}^{1}\).

Evaluate the integral

Substitute the limits to the result of last step to calculate the numerical value of the double integral: \([\frac{1}{6}\cdot1^{6} - \frac{1}{7}\cdot1^{7}] - [\frac{1}{6}\cdot0^{6} - \frac{1}{7}\cdot0^{7}] = \frac{1}{6} - \frac{1}{7} = \frac{1}{42}.\