Evaluate the following integral using a change of variables.

Integral from 0 to 4∫04Integral from y to y plus 6∫yy+6StartRoot x minus y EndRootx−y 

dxdy

To evaluate the integral 

\[
\int_{0}^{4} \int_{y}^{y+6} \sqrt{x - y} \, dx \, dy
\]

we’ll proceed by using a change of variables. The square root term, \( \sqrt{x - y} \), suggests the substitution \( u = x - y \).

### Step 1: Define the Substitution
Let
\[
u = x - y.
\]
Then,
\[
x = u + y.
\]

Taking the derivative of \( u \) with respect to \( x \), we get
\[
\frac{du}{dx} = 1 \Rightarrow dx = du.
\]

### Step 2: Update the Limits for \( u \)
For fixed \( y \), we have:
- When \( x = y \), then \( u = y - y = 0 \).
- When \( x = y + 6 \), then \( u = (y + 6) - y = 6 \).

Thus, the integral with respect to \( x \) becomes an integral with respect to \( u \) from \( 0 \) to \( 6 \).

### Step 3: Rewrite the Integral
Substituting \( u = x - y \) and \( dx = du \), we get:
\[
\int_{0}^{4} \int_{0}^{6} \sqrt{u} \, du \, dy.
\]

### Step 4: Evaluate the Inner Integral
The inner integral is with respect to \( u \):
\[
\int_{0}^{6} \sqrt{u} \, du.
\]

Recall that \( \sqrt{u} = u^{1/2} \). So,
\[
\int_{0}^{6} u^{1/2} \, du = \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{6} = \frac{2}{3} u^{3/2} \Big|_{0}^{6}.
\]

Evaluating at the limits,
\[
= \frac{2}{3} \left(6^{3/2} - 0^{3/2}\right) = \frac{2}{3} \cdot 6^{3/2}.
\]

Since \( 6^{3/2} = 6 \cdot \sqrt{6} \),
\[
= \frac{2}{3} \cdot 6 \sqrt{6} = 4 \sqrt{6}.
\]

### Step 5: Evaluate the Outer Integral
Now we have reduced the original integral to
\[
\int_{0}^{4} 4 \sqrt{6} \, dy.
\]

This is simply a constant with respect to \( y \), so we get:
\[
4 \sqrt{6} \int_{0}^{4} dy = 4 \sqrt{6} \cdot y \Big|_{0}^{4} = 4 \sqrt{6} \cdot 4 = 16 \sqrt{6}.
\]

### Final Answer
\[
\int_{0}^{4} \int_{y}^{y+6} \sqrt{x - y} \, dx \, dy = 16 \sqrt{6}.
\]

Would you like to see more details on any step?

### Related Questions

1. How would the evaluation change if the upper bound in the outer integral was modified?
2. What would happen if the square root term in the integrand was different, such as \( \sqrt{x + y} \)?
3. How can the method of substitution be applied to other integrals involving square roots?
4. What would the answer be if we integrated over a different region?
5. How can this integral be interpreted geometrically?

### Tip
When using substitution for integrals, remember to carefully adjust the limits of integration for the new variable to ensure accurate results.

Evaluate the following integral using a change of variables: Integral from 0 to 4 of the integral from y to y+6 of √(x - y) dx dy

Learn how to evaluate a double integral from 0 to 4 of the integral from y to y+6 of √(x - y) using a change of variables. This solution applies substitution and covers steps for transforming limits, making it suitable for college-level calculus.

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