Consider the region bounded by 𝑦 = 𝑥 ^2 + 5, 𝑦 = 0, 𝑥 = 0, 𝑥 = 3 
a) Compute the area of the region using dy and the horizontal method

To compute the area of the region bounded by the curves using the horizontal method (with respect to \(y\)), we need to express \(x\) in terms of \(y\), since we will integrate horizontally along the \(y\)-axis.

### Step 1: Rearranging the equation for \(x\)
The given equation is:
\[
y = x^2 + 5
\]
To express \(x\) in terms of \(y\), solve for \(x\):
\[
x = \sqrt{y - 5}
\]

### Step 2: Limits of integration
Next, identify the limits of integration for \(y\). The region is bounded vertically by \(y = 0\) and the curve. The curve intersects the line \(x = 3\), so substitute \(x = 3\) into the equation \(y = x^2 + 5\):
\[
y = 3^2 + 5 = 9 + 5 = 14
\]
Thus, the region extends from \(y = 5\) to \(y = 14\).

### Step 3: Setting up the integral
Since we are using the horizontal method (with respect to \(y\)), we will integrate along the \(y\)-axis. The width of each horizontal strip is given by the distance between \(x = 0\) and \(x = \sqrt{y - 5}\). The area of a small horizontal strip is the width (\( \sqrt{y - 5} \)) multiplied by the small change in height \(dy\).

The total area \(A\) is:
\[
A = \int_{y = 5}^{y = 14} \sqrt{y - 5} \, dy
\]

### Step 4: Solving the integral
To compute the integral:
\[
\int \sqrt{y - 5} \, dy
\]
Let \( u = y - 5 \), so \( du = dy \). The limits of integration change accordingly: when \( y = 5 \), \( u = 0 \); when \( y = 14 \), \( u = 9 \).

Thus, the integral becomes:
\[
A = \int_{u = 0}^{u = 9} \sqrt{u} \, du
\]

Now integrate:
\[
A = \int_{u = 0}^{u = 9} u^{1/2} \, du = \left[\frac{2}{3} u^{3/2}\right]_{0}^{9}
\]
\[
A = \frac{2}{3} \left[9^{3/2} - 0^{3/2}\right] = \frac{2}{3} \times 27 = 18
\]

### Final Answer:
The area of the region is \( \boxed{18} \) square units.

Do you need further details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How would the area change if the upper bound for \(x\) were \(x = 4\)?
2. What if the curve were \(y = x^2 + 6\) instead of \(y = x^2 + 5\)?
3. How would the setup differ if using vertical strips instead of horizontal?
4. How do we determine the limits of integration for more complex curves?
5. What is the geometric significance of solving integrals using \(dy\)?

**Tip:** Always check if switching to \(dy\) simplifies the integration when the curve is easier to express in terms of \(y\).

Consider the region bounded by 𝑦 = 𝑥^2 + 5, 𝑦 = 0, 𝑥 = 0, 𝑥 = 3. a) Compute the area of the region using dy and the horizontal method.

Learn how to compute the area of the region bounded by the curves y = x^2 + 5, y = 0, x = 0, and x = 3 using the horizontal strip method (dy). This problem covers integral calculus and involves finding the limits of integration, applying u-substitution, and solving a definite integral. Suitable for undergraduate calculus students.

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