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Flashcards: Find Cross-Sections: Squares & Rectangles

Find the volume of a pyramid whose base is a square with sides of length and whose height is .

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Flashcards: Find Cross-Sections: Squares & Rectangles

Find the volume of a pyramid whose base is a square with sides of length and whose height is .

Your answer:

Correct answer:

Explanation:

First, it is important to consider the shape of this solid. This solid is a pyramid, with one square face and four triangular faces. Through a relationship of similar triangles, we are able to relate the known information (a height of and a base side length of ) to our general variables for side length (x) and height of the pyramid (y) . We can think of this plotted on the coordinate plane, with the width of the pyramid solid being in the direction.

$\frac{x}{3}=\frac{y}{10} \Rightarrow x=\frac{3y}{10}$

Since the side length can be squared to find a general formula for the area of the pyramid’s base (A=x^2) , this can be applied to the volume using cross-sections formula $(V=\int_{a}^{b}A(x) dx)$ as the next step. Note that the equation $x=\frac{3y}{10}$ solved for above is in terms of . Our new volume function, therefore, is also in terms of $y: V=\int_{a}^{b}(\frac{3y}{10})^2 \: dy$

Because the pyramid reaches a maximum height of , and we assume the pyramid’s starting height is at y=0 , the appropriate bounds for the integral expression are: $0\leq y\leq 10$ .

To wrap up this problem, combine the above information into one cohesive expression:

$V=\int_{0}^{10}(\frac{3y}{10})^2 dy =\frac{9}{100} \int_{0}^{10}y^2dy =\frac{9}{100}[\frac{1}{3}y^3] |_0^{10}=\frac{9}{100}*\frac{1}{3}*1000=30$

Flashcards: Find Cross-Sections: Squares & Rectangles

Find the volume of the solid whose base is bounded by y=x+4 and y=x^2-2 and whose cross-sections are rectangles of height and perpendicular to the axis.

50.5

62.5

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Flashcards: Find Cross-Sections: Squares & Rectangles

Find the volume of the solid whose base is bounded by y=x+4 and y=x^2-2 and whose cross-sections are rectangles of height and perpendicular to the axis.

Your answer:

50.5

Your answer:

Correct answer:

62.5

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a rectangle is A=l*w . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) dx)$ , we get the following: $V=\int_{a}^{b}lw \: dx$ .

Next, an expression for must be determined. The problem specifies the length (or height) of the rectangle cross-sections is . This just leaves the value of w to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between y=x+4 and y=x^2-2 , therefore, w=(x+4)-(x^2-2)=6+x-x^2 .

Since the region is bounded by y=x+4 and y=x^2-2 , the base has the following domain: $-2\leq x\leq 3$ .

Putting this all together, we find the following:

$V=\int_{-2}^{3}(3)(6+x-x^2) dx$ $= 3[6x+\frac{1}{2}x^2-\frac{1}{3}x^3] |_{-2}^3=3(18+\frac{9}{2}-9+12-2-\frac{8}{3})=62.5$

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All Calculus AB Resources

Flashcards: Find Cross-Sections: Squares & Rectangles

Flashcards: Find Cross-Sections: Squares & Rectangles

Flashcards: Find Cross-Sections: Squares & Rectangles

Flashcards: Find Cross-Sections: Squares & Rectangles

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Flashcards: Find Cross-Sections: Squares & Rectangles

Flashcards: Find Cross-Sections: Squares & Rectangles

Flashcards: Find Cross-Sections: Squares & Rectangles

Flashcards: Find Cross-Sections: Squares & Rectangles

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DMCA Complaint

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