Problem 8 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Evaluate the double integrals.\(\int_{0}^{1} \int_{1}^{5} y \sqrt{1-y^{2}} d x d y\)

Step-by-Step Solution

Answer

The value of the double integral is \frac{4}{3}\.

Step 1: Setup the integral

Evaluate the inner integral first. Given the double integral \(\textstyle \int_{0}^{1} \int_{1}^{5} y \sqrt{1-y^{2}} \ d x \ d y\), this means integrating the function \(\textstyle y \sqrt{1-y^{2}}\) with respect to \(\textstyle x\) first and then \(\textstyle y\).

Step 2: Integrate the inner integral

Since \(\textstyle x\) is from 1 to 5 and the integrand \(\textstyle y \sqrt{1-y^{2}}\) does not depend on \(\textstyle x\), treat \(\textstyle y \sqrt{1-y^{2}}\) as a constant with respect to \(\textstyle x\). Thus the inner integral evaluates as: \(\textstyle \int_{1}^{5} y \sqrt{1-y^{2}} \ d x = y \sqrt{1-y^{2}} \int_{1}^{5} \ d x = y \sqrt{1-y^{2}} (5 - 1) = 4 y \sqrt{1-y^{2}}\).

Step 3: Setup the resulting single integral

After computing the inner integral, the double integral reduces to a single integral: \(\textstyle\int_{0}^{1} 4 y \sqrt{1-y^{2}} \ d y\).

Step 4: Evaluate the single integral

Use the substitution method to solve the single integral. Let \(\textstyle u = 1 - y^{2}\), then \(\textstyle du = -2y \ dy\) or \(\textstyle y \ dy = - \frac{1}{2} \ du\). Also, change the limits of integration: when \(\textstyle y = 0\), \(\textstyle u = 1\) and when \(\textstyle y = 1\), \(\textstyle u = 0\). Thus the integral becomes: \(\textstyle \int_{1}^{0} 4 (-\frac{1}{2}) \sqrt{u} \ du = -2 \int_{1}^{0} \sqrt{u} \ du = 2 \int_{0}^{1} u^{1/2} \ du\).

Step 5: Integrate and solve

Evaluate \(\textstyle 2 \int_{0}^{1} u^{1/2} \ du\): \(\textstyle 2 \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = 2 \left( \frac{2}{3} (1^{3/2} - 0^{3/2}) \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3}\).

Key Concepts

Substitution Method

The substitution method is critical for simplifying complex integrals, especially in double integrals. It involves changing variables to make integration easier. In our problem, we used the substitution \( u = 1 - y^2 \). By this substitution, we transformed the integrand into a simpler form. Here’s how it works:

• First, identify the substitution: let \( u = 1 - y^2 \).
• Compute the differential: \( du = -2y \, dy \) or \( y \, dy = -\frac{1}{2} \, du \).
• Change the limits of integration accordingly. When \( y = 0 \), \( u = 1 \); when \( y = 1 \), \( u = 0 \).
• Rewrite the integral in terms of \( u \). This simplification often results in an easier integral to evaluate.

Inner Integral

The inner integral is the first integration you perform when evaluating a double integral. For our problem, the given double integral is \( \textstyle \ int_{0}^{1} \ int_{1}^{5} y \ sqrt{1-y^{2}} \, dx \, dy \).

In this step, we integrate with respect to \( x \), treating \( y \ sqrt{1-y^{2}} \) as a constant because it doesn’t contain \( x \).

• Since \( y \ sqrt{1-y^{2}} \) is independent of \( x \), the inner integral can be simplified to be \( \textstyle \ int_{1}^{5} \, dx \).
• The result of this inner integration is then multiplied by \( y \ sqrt{1-y^{2}} \).

Using this technique, we simplified the integral to \( 4y \ sqrt{1-y^{2}} \).

Single Integral

Once you calculate the inner integral, the double integral reduces to a single integral. For our example, we get: \( \textstyle \ int_{0}^{1} 4y \ sqrt{1-y^{2}} \, dy \).
This simplification makes the problem easier to solve using standard integration techniques.

Now, instead of evaluating a complex double integral, you only need to handle a single integral.

• The process includes rewriting our integral in a simpler form, often using techniques like substitution.
• Our single integral was solved by substituting \( u = 1 - y^2 \), turning it into \( \textstyle 2 \ int_{0}^{1} u^{1/2} \, du \).

This approach is often much more manageable.

Integration Limits

Integration limits define the range over which you integrate. For double integrals, you have limits for both inner and outer integrals. In our problem, we had:

• The inner integral limits from 1 to 5 for \( x \).
• The outer integral limits from 0 to 1 for \( y \).

When applying substitution, you must also change these limits accordingly.
Example:

• Original limits for \( y \) were 0 to 1.
• After substituting \( u = 1 - y^2 \), new limits for \( u \) became 1 to 0.
• We then reversed the integration order, which changes the integral from \( \ int_{1}^{0} \) to \( \ int_{0}^{1} \).

This ensures the integral remains accurate and solvable.