AP Calculus BC 8.7 Volumes with Cross Sections: Squares and Rectangles Study Notes - New Syllabus
AP Calculus BC 8.7 Volumes with Cross Sections: Squares and Rectangles Study Notes- New syllabus
AP Calculus BC 8.7 Volumes with Cross Sections: Squares and Rectangles Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- Volumes with Cross Sections: Squares and Rectangles
Volumes with Cross Sections: Squares and Rectangles
Volumes with Cross Sections: Squares
When the base of a solid lies in a region of the \( xy \)-plane and the cross sections perpendicular to an axis are squares, the volume can be computed by integrating the area of each square slice.
- The side length of each square cross section is determined by the width of the base at that position.
- If slices are perpendicular to the x-axis: \( s(x) = f_{\text{top}}(x) – f_{\text{bottom}}(x) \)
- If slices are perpendicular to the y-axis: \( s(y) = f_{\text{right}}(y) – f_{\text{left}}(y) \)
Volume Formulas:
If perpendicular to x-axis: \( V = \displaystyle \int_{a}^{b} [s(x)]^2 \, dx \)
If perpendicular to y-axis: \( V = \displaystyle \int_{c}^{d} [s(y)]^2 \, dy \)
Procedure:
- Identify the base region and draw a sketch.
- Determine the limits of integration based on the given axis of cross sections.
- Find the formula for the side length of the square at each slice.
- Square the side length to get the area of the cross section.
- Integrate to find the volume.
Example:
The base of a solid is the region between \( y = \sqrt{x} \) and \( y = 0 \) for \( 0 \leq x \leq 4 \). Cross sections perpendicular to the x-axis are squares. Find the volume of the solid.
▶️ Answer/Explanation
Side length: \( s(x) = \sqrt{x} – 0 = \sqrt{x} \)
Volume:
\( V = \displaystyle \int_{0}^{4} (\sqrt{x})^2 \, dx = \displaystyle \int_{0}^{4} x \, dx \)
\( = \left[ \dfrac{x^2}{2} \right]_{0}^{4} = \dfrac{16}{2} = 8 \)
Final Answer: \( 8 \) cubic units
Example:
The base is the region bounded by \( x = y^2 \) and \( x = 4 \) for \( 0 \leq y \leq 2 \). Cross sections perpendicular to the y-axis are squares. Find the volume.
▶️ Answer/Explanation
Side length: \( s(y) = 4 – y^2 \)
Volume: \( V = \displaystyle \int_{0}^{2} (4 – y^2)^2 \, dy \)
Expand: \( V = \displaystyle \int_{0}^{2} (16 – 8y^2 + y^4) \, dy \)
Integrate: \( V = \left[ 16y – \dfrac{8y^3}{3} + \dfrac{y^5}{5} \right]_{0}^{2} \)
Substitute: \( V = 32 – \dfrac{64}{3} + \dfrac{32}{5} \)
Common denominator \( 15 \): \( V = \dfrac{480 – 320 + 96}{15} = \dfrac{256}{15} \)
Final Answer: \( \dfrac{256}{15} \) cubic units
Example:
The base is the region between \( y = \sqrt{x} \) and \( y = x^2 \) for \( 0 \leq x \leq 1 \). Cross sections perpendicular to the x-axis are squares. Find the volume.
▶️ Answer/Explanation
Side length: \( s(x) = \sqrt{x} – x^2 \)
Volume: \( V = \displaystyle \int_{0}^{1} (\sqrt{x} – x^2)^2 \, dx \)
Expand: \( V = \displaystyle \int_{0}^{1} (x – 2x^{\frac{5}{2}} + x^4) \, dx \)
Integrate: \( V = \left[ \dfrac{x^2}{2} – \dfrac{4x^{\frac{7}{2}}}{7} + \dfrac{x^5}{5} \right]_{0}^{1} \)
Substitute: \( V = \dfrac{1}{2} – \dfrac{4}{7} + \dfrac{1}{5} \)
Common denominator \( 70 \): \( V = \dfrac{35 – 40 + 14}{70} = \dfrac{9}{70} \)
Final Answer: \( \dfrac{9}{70} \) cubic units
Volumes with Cross Sections: Rectangles
When the base of a solid lies in the \( xy \)-plane and cross sections perpendicular to the \( x \)-axis (or \( y \)-axis) are rectangles, the volume can be determined by integrating the area of each cross section along the given interval.
The height of each rectangle is proportional to the distance between the two bounding curves of the base. If the height is given as a constant multiple \( k \) of the base length, then:
\( A(x) = k \cdot \text{Base Length} \)
For perpendicular cross sections to the \( x \)-axis:
\( V = \displaystyle \int_{a}^{b} k \left[ f(x) – g(x) \right] \, dx \)
For perpendicular cross sections to the \( y \)-axis:
\( V = \displaystyle \int_{c}^{d} k \left[ f(y) – g(y) \right] \, dy \)
Example:
The base of a solid is the region bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\). The cross sections perpendicular to the \(x\)-axis are rectangles whose height is twice their base (length in the \(y\)-direction).
▶️ Answer/Explanation
Here, the base length \(b(x)\) is given by \(y = \sqrt{x}\), so \(b(x) = \sqrt{x}\).
Height \(h(x) = 2b(x) = 2\sqrt{x}\).
Area of each cross section: \(A(x) = b(x) \cdot h(x) = \sqrt{x} \cdot 2\sqrt{x} = 2x\).
Volume: \(V = \displaystyle \int_{0}^{4} 2x \, dx = \left[ x^{2} \right]_{0}^{4} = 16\) cubic units.
Example:
The base of a solid is bounded by \(y = x^{2}\) and \(y = 4\). The cross sections perpendicular to the \(x\)-axis are rectangles of constant height \(h = 3\).
▶️ Answer/Explanation
At each \(x\), the base length is \(b(x) = 4 – x^{2}\).
Height is constant: \(h(x) = 3\).
Area: \(A(x) = b(x) \cdot h(x) = 3(4 – x^{2})\).
Volume: \(V = \displaystyle \int_{-2}^{2} 3(4 – x^{2}) \, dx = 3 \left[ 4x – \dfrac{x^{3}}{3} \right]_{-2}^{2} = 3 \left[ (8 – \dfrac{8}{3}) – (-8 + \dfrac{8}{3}) \right] = 3 \left[ \dfrac{16}{3} + \dfrac{16}{3} \right] = 3 \cdot \dfrac{32}{3} = 32\) cubic units.
Example:
The base of a solid is the region under \(y = \sqrt{4 – x^{2}}\) from \(x = -2\) to \(x = 2\). The cross sections perpendicular to the \(x\)-axis are rectangles whose height is equal to the square of their base.
▶️ Answer/Explanation
Base length: \(b(x) = 2\sqrt{4 – x^{2}}\).
Height: \(h(x) = [b(x)]^{2} = [2\sqrt{4 – x^{2}}]^{2} = 4(4 – x^{2})\).
Area: \(A(x) = b(x) \cdot h(x) = 2\sqrt{4 – x^{2}} \cdot 4(4 – x^{2}) = 8(4 – x^{2})^{3/2}\).
Volume: \(V = \displaystyle \int_{-2}^{2} 8(4 – x^{2})^{3/2} \, dx\).
Using symmetry: \(V = 2 \displaystyle \int_{0}^{2} 8(4 – x^{2})^{3/2} \, dx\).
Let \(x = 2\sin\theta\), then \(dx = 2\cos\theta \, d\theta\), and \(4 – x^{2} = 4\cos^{2}\theta\).
Volume: \(V = 16 \displaystyle \int_{0}^{\pi/2} (4\cos^{2}\theta)^{3/2} \cdot 2\cos\theta \, d\theta = 32 \displaystyle \int_{0}^{\pi/2} 8\cos^{4}\theta \, d\theta\).
\(\displaystyle \int_{0}^{\pi/2} \cos^{4}\theta \, d\theta = \dfrac{3\pi}{16} \cdot \dfrac{1}{2}\) using reduction formulas.
Final: \(V = 32 \cdot \dfrac{3\pi}{8} = 12\pi\) cubic units.