To find the value of \( a \) for which the area between the curves \( y^2 = 4ax \) and \( x^2 = 4ay \) is 1 unit, we will follow these steps:
### Step 1: Identify the curves
The first curve is given by \( y^2 = 4ax \), which is a rightward-opening parabola. The second curve is given by \( x^2 = 4ay \), which is an upward-opening parabola.
### Step 2: Find the points of intersection
To find the points of intersection, we will solve the equations simultaneously. From the first equation, we can express \( x \) in terms of \( y \):
\[
x = \frac{y^2}{4a}
\]
Now, substituting this expression for \( x \) into the second equation:
\[
\left(\frac{y^2}{4a}\right)^2 = 4ay
\]
This simplifies to:
\[
\frac{y^4}{16a^2} = 4ay
\]
Multiplying both sides by \( 16a^2 \) gives:
\[
y^4 = 64a^3y
\]
Assuming \( y \neq 0 \), we can divide both sides by \( y \):
\[
y^3 = 64a^3
\]
Taking the cube root of both sides:
\[
y = 4a
\]
Now substituting \( y = 4a \) back into the expression for \( x \):
\[
x = \frac{(4a)^2}{4a} = 4a
\]
Thus, the points of intersection are \( (4a, 4a) \) and the origin \( (0, 0) \).
### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 4a \) can be expressed as:
\[
A = \int_{0}^{4a} \left( y_{\text{upper}} - y_{\text{lower}} \right) \, dx
\]
From the first equation, we find \( y_{\text{upper}} = 2\sqrt{a\sqrt{x}} \) and from the second equation, \( y_{\text{lower}} = \frac{x^2}{4a} \).
Thus, the area becomes:
\[
A = \int_{0}^{4a} \left( 2\sqrt{a}\sqrt{x} - \frac{x^2}{4a} \right) \, dx
\]
### Step 4: Evaluate the integral
Now we compute the integral:
\[
A = \int_{0}^{4a} 2\sqrt{a}\sqrt{x} \, dx - \int_{0}^{4a} \frac{x^2}{4a} \, dx
\]
Calculating the first integral:
\[
\int 2\sqrt{a}\sqrt{x} \, dx = 2\sqrt{a} \cdot \frac{2}{3} x^{3/2} = \frac{4\sqrt{a}}{3} x^{3/2}
\]
Evaluating from \( 0 \) to \( 4a \):
\[
\left[ \frac{4\sqrt{a}}{3} (4a)^{3/2} \right] = \frac{4\sqrt{a}}{3} \cdot 8a^{3/2} = \frac{32a^2}{3}
\]
Calculating the second integral:
\[
\int \frac{x^2}{4a} \, dx = \frac{1}{4a} \cdot \frac{x^3}{3} = \frac{x^3}{12a}
\]
Evaluating from \( 0 \) to \( 4a \):
\[
\left[ \frac{(4a)^3}{12a} \right] = \frac{64a^2}{12} = \frac{16a^2}{3}
\]
### Step 5: Combine the results
Now, substituting back into the area formula:
\[
A = \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3}
\]
### Step 6: Set the area equal to 1
We set the area equal to 1:
\[
\frac{16a^2}{3} = 1
\]
Solving for \( a^2 \):
\[
16a^2 = 3 \implies a^2 = \frac{3}{16} \implies a = \frac{\sqrt{3}}{4}
\]
### Final Answer
Thus, the value of \( a \) for which the area between the curves is 1 unit is:
\[
\boxed{\frac{\sqrt{3}}{4}}
\]
