change the order of integration and evaluate double integral x2 dydx from 0 to b/aroot a2-x2 and 0 to a

We are given the double integral:  

\[
I = \int_0^a \int_0^{\frac{b}{\sqrt{a^2 - x^2}}} x^2 \, dy \, dx
\]

### **Step 1: Understanding the Region of Integration**
The limits describe a region where:  
- \( x \) varies from \( 0 \) to \( a \).  
- \( y \) varies from \( 0 \) to \( \frac{b}{\sqrt{a^2 - x^2}} \).  

To change the order of integration, we express the region in terms of \( y \) first.  
From the upper limit of \( y \), we solve for \( x \):  

\[
y = \frac{b}{\sqrt{a^2 - x^2}}
\]

Squaring both sides:

\[
y^2 (a^2 - x^2) = b^2
\]

Rearrange:

\[
x^2 = a^2 - \frac{b^2}{y^2}
\]

So, for a fixed \( y \), \( x \) varies from \( 0 \) to \( \sqrt{a^2 - \frac{b^2}{y^2}} \). The value of \( y \) starts from \( 0 \) and extends to its maximum when \( x = 0 \), which gives \( y = \frac{b}{a} \).  

### **Step 2: Writing the Integral with Changed Order**
The new limits are:  
- \( y \) varies from \( 0 \) to \( \frac{b}{a} \).  
- \( x \) varies from \( 0 \) to \( \sqrt{a^2 - \frac{b^2}{y^2}} \).  

Thus, the integral becomes:

\[
I = \int_0^{\frac{b}{a}} \int_0^{\sqrt{a^2 - \frac{b^2}{y^2}}} x^2 \, dx \, dy
\]

### **Step 3: Evaluating the Inner Integral**
We compute:

\[
\int_0^{\sqrt{a^2 - \frac{b^2}{y^2}}} x^2 \, dx
\]

Using the standard formula:

\[
\int x^2 \, dx = \frac{x^3}{3}
\]

Evaluating from \( 0 \) to \( \sqrt{a^2 - \frac{b^2}{y^2}} \):

\[
\frac{1}{3} \left( a^2 - \frac{b^2}{y^2} \right)^{\frac{3}{2}}
\]

Thus, our integral reduces to:

\[
I = \int_0^{\frac{b}{a}} \frac{1}{3} \left( a^2 - \frac{b^2}{y^2} \right)^{\frac{3}{2}} dy
\]

### **Step 4: Substituting \( z = \frac{b}{y} \)**
Define \( z = \frac{b}{y} \), so \( dz = -\frac{b}{y^2} dy \), or equivalently:

\[
dy = -\frac{b}{z^2} dz
\]

When \( y = 0 \), \( z \to \infty \), and when \( y = \frac{b}{a} \), \( z = a \). Thus, the limits become \( z \) from \( \infty \) to \( a \), and our integral transforms into:

\[
I = \int_{\infty}^{a} \frac{1}{3} \left( a^2 - z^2 \right)^{\frac{3}{2}} \cdot \left( -\frac{b}{z^2} \right) dz
\]

Since the negative sign flips the limits, we write:

\[
I = \frac{b}{3} \int_a^{\infty} \frac{\left( a^2 - z^2 \right)^{\frac{3}{2}}}{z^2} dz
\]

This integral is more challenging and requires special functions (such as Beta functions) to evaluate explicitly.

Would you like a numerical approach or further simplifications?  

### **Relative Questions**
1. How do we interpret changing the order of integration geometrically?  
2. Can we solve this integral using polar coordinates?  
3. What are some techniques to evaluate difficult integrals involving square roots?  
4. How does substitution help in integral evaluation?  
5. How do Beta and Gamma functions relate to this integral?  

### **Tip**
When changing the order of integration, always sketch the region first to understand the new limits properly.

Change the order of integration and evaluate the double integral x^2 dydx from 0 to b/√(a^2 - x^2) and 0 to a.

Learn how to change the order of integration and evaluate the double integral ∫∫ x^2 dydx for given limits. This problem covers core calculus concepts such as multiple integrals, substitution methods, and definite integration techniques. Suitable for undergraduate students studying calculus.

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