Problem 18 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Evaluate the double integrals.\(\int_{0}^{3} \int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x y d x d y\)

Step-by-Step Solution

Answer

8.578125

Step 1: Understand the Integration Limits

The given double integral \(\int_{0}^{3} \int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x y d x d y\) has the limits for x as \[ \frac{y^{2}}{4} \ \text{ and } \sqrt{10-y^{2}} \] and for y from 0 to 3.

Step 2: Integrate with respect to x

First, perform the integration with respect to x. The inner integral is \(\int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x y d x \). Since y is treated as a constant here, the integral becomes: \[ y \int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x d x \] This evaluates to: \[ \left[ \frac{x^{2} y}{2} \right]_{y^{2} / 4}^{\sqrt{10-y^{2}}} \]

Step 3: Substitute the limits for x

Substitute the upper and lower limits for x into the integrated result: \[\left( \frac{(\sqrt{10-y^{2}})^{2} y}{2} \right) \ - \ \left( \frac{\left( \frac{y^{2}}{4} \right)^{2} y}{2} \right) = \left( \frac{10-y^{2}}{2} y \right) \ - \ \left( \frac{y^{5}}{32} \right) \]

Step 4: Simplify the integrand

Simplify the expression inside the integral: \[\frac{10 y - y^{3}}{2} - \frac{y^{5}}{32}\]

Step 5: Integrate with respect to y

Now, evaluate the outer integral from 0 to 3: \[\begin{aligned} \int_{0}^{3} \left( \frac{10 y - y^{3}}{2} - \frac{y^{5}}{32} \right) d y \end{aligned} \]

Step 6: Calculate the integral

Evaluate each term separately: \[\begin{aligned} \int_{0}^{3} \frac{10 y}{2} d y &= 5 \left[ \frac{y^{2}}{2} \right]_{0}^{3} = 5 \frac{9}{2} = 22.5 \ \ \int_{0}^{3} \frac{y^{3}}{2} d y &= \frac{1}{2} \left[ \frac{y^{4}}{4} \right]_{0}^{3} = \frac{1}{2} \frac{81}{4} = \frac{81}{8} = 10.125 \ \ \int_{0}^{3} \frac{1}{32} y^{5} d y &= \frac{1}{32} \left[ \frac{y^{6}}{6} \right]_{0}^{3} = \frac{1}{32} \frac{729}{6} = \frac{729}{192} = 3.796875 \ \ \end{aligned} \]

Step 7: Combine the results

Combine these results and subtract: \[\begin{aligned} 22.5 - 10.125 - 3.796875 = 8.578125 \end{aligned} \]

Key Concepts

Understanding Integration Limits

In double integrals, the integration limits define the region over which we are integrating. For the given integral: \( \int_{0}^{3} \int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x y d x d y\), the limits for \text{x} are \[\frac{y^{2}}{4} \text{ and } \sqrt{10-y^{2}} \] and for \text{y} from 0 to 3.
\text{y} varies first from 0 to 3, setting the vertical bound of the region, while \text{x} varies based on the current value of \text{y}, from \[\frac{y^{2}}{4} \] to \[\sqrt{10-y^{2}}\].
Visualizing these limits helps understand the shape and size of the integration region.

Integrating with Respect to x

To integrate with respect to \text{x}, we keep \text{y} constant. The inner integral is \int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x y d x. Here, \text{y} is treated as a constant.
The integration simplifies as follows: \text{y} \int_{y^{2} / 4}^{\sqrt{10-y^{2}}} x d x.

Inside this integral, we only need to focus on \int x d x, which is \[ \frac{x^{2}}{2}\]. So, we get: \left[ \frac{x^{2} y}{2} \right]_{y^{2} / 4}^{\sqrt{10-y^{2}}}.

Substitute the Limits and Simplify

Now we substitute the integration limits for \text{x} into our expression: \[\left( \frac{\left( \sqrt{10-y^{2}} \right)^{2} y}{2} \right) - \left( \frac{\left( \frac{y^{2}}{4} \right)^{2} y}{2} \right)\].
This will expand and simplify to: \[\left( \frac{10 - y^{2}}{2} y \right) - \left( \frac{y^{5}}{32} \right)\].

The resulting expression inside the integral now is: \frac{10 y - y^{3}}{2} - \frac{y^{5}}{32}.

Integrating with Respect to y and Evaluating the Integral

Lastly, we integrate with respect to \text{y}. The integrand is \int_{0}^{3} \left( \frac{10 y - y^{3}}{2} - \frac{y^{5}}{32} \right) d y.
Breaking down into parts and integrating each term separately gives:

• \[ \int_{0}^{3} \frac{10 y}{2} d y = 5 \left[ \frac{y^{2}}{2} \right]_{0}^{3} = 5 \frac{9}{2} = 22.5\]
  
• \[\int_{0}^{3} \frac{y^{3}}{2} d y = \frac{1}{2} \left[ \frac{y^{4}}{4} \right]_{0}^{3}= \frac{1}{2} \frac{81}{4} = \frac{81}{8} = 10.125\]
  
• \[ \int_{0}^{3} \frac{1}{32} y^{5} d y = \frac{1}{32} \left[ \frac{y^{6}}{6} \right]_{0}^{3} = \frac{1}{32} \frac{729}{6} = \frac{729}{192} = 3.796875\]
  

Finally, combine the results: \[22.5 - 10.125 - 3.796875 = 8.578125\].
This is the value of the original double integral.