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Flashcards: Find Cross-Sections: Triangles & Semicircles

Find the volume of the solid whose cross-sections are equilateral triangles and whose base is a disk of radius .

$V=\int_{-R}^{R}\sqrt{3} (R-x^2) dx$

$V=\int_{0}^{R}\sqrt{3} (R^2-x^2) dx$

$V=\int_{-R}^{R}\sqrt{3} (R^2-x^2) dx$

$V=\int_{-R}^{R}(R^2-x^2) dx$

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Flashcards: Find Cross-Sections: Triangles & Semicircles

Find the volume of the solid whose cross-sections are equilateral triangles and whose base is a disk of radius .

Your answer:

$V=\int_{-R}^{R}\sqrt{3} (R-x^2) dx$

Your answer:

$V=\int_{0}^{R}\sqrt{3} (R^2-x^2) dx$

Your answer:

$V=\int_{-R}^{R}(R^2-x^2) dx$

Correct answer:

$V=\int_{-R}^{R}\sqrt{3} (R^2-x^2) dx$

Explanation:

Because the disk is of radius R, the base is defined by the following formula: x^2+y^2=R^2 .

The correct formula for the area of an equilateral triangle is as follows:

$A=\frac{\sqrt{3}}{4}s^2$ , with being the side length of the triangle.

By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) dx)$ , we get the following: $V=\int_{a}^{b}\frac{\sqrt{3}}{4}s^2 dx$ .

The radius R defines the bounds as being [-R,R] . Next, s can be found by understanding that the value is the distance from the top to the bottom of the circle at any given point along [-R,R] . The length of one side of the equilateral triangle, therefore, is $s=2\sqrt{R^2-x^2}$ .

Putting it all together, the following is obtained:

$V=\int_{-R}^{R}\frac{3}{4}(2\sqrt{R^2-x^2})^2 dx=\int_{-R}^{R}\sqrt{3 }(R^2-x^2) dx$

*Note: the problem did not specify if the cross sections were perpendicular to the or axis. Because the base is a circle, this should not change the resulting volume. The only difference should be the use of or as variables in the correct expression.

Flashcards: Find Cross-Sections: Triangles & Semicircles

Find the volume of the solid whose base is bounded by the circle x^2+y^2=4 and whose cross-sections are right isosceles triangles perpendicular to the axis, with one leg on the base of the solid.

$\frac{64}{5}$

$\frac{64}{3}$

$\frac{56}{3}$

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Flashcards: Find Cross-Sections: Triangles & Semicircles

Find the volume of the solid whose base is bounded by the circle x^2+y^2=4 and whose cross-sections are right isosceles triangles perpendicular to the axis, with one leg on the base of the solid.

Your answer:

$\frac{64}{5}$

Your answer:

$\frac{56}{3}$

Your answer:

Correct answer:

$\frac{64}{3}$

Explanation:

Because the base is a circle of radius , the bounds are defined as [-2,2] .

The area of a right isosceles triangle can be found using the formula $A=\frac{1}{2}b^2$ , where is the leg length of the triangle. By applying this to our general volume formula $(V=\int_{a}^{b}A(x) dx)$ , we get the following: $V=\int_{a}^{b}\frac{1}{2}b^2 dx$ .

The expression for can be found by understanding the fact that the leg of the triangle is on the base of the solid. The value is twice the height of the semicircle $y=\sqrt{4-x^2}$ . $b=2\sqrt{4-x^2}$

Putting it all together, the following is obtained:

$V=\int_{-2}^{2}\frac{1}{2}(2\sqrt{4-x^2})^2 dx =\int_{-2}^{2}8-2x2 dx$ $=[8x-\frac{2}{3}x^3]|_{-2}^2=16-\frac{16}{3}+16-\frac{16}{3}=\frac{64}{3}$

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All Calculus AB Resources

Flashcards: Find Cross-Sections: Triangles & Semicircles

Flashcards: Find Cross-Sections: Triangles & Semicircles

Flashcards: Find Cross-Sections: Triangles & Semicircles

Flashcards: Find Cross-Sections: Triangles & Semicircles

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Flashcards: Find Cross-Sections: Triangles & Semicircles

Flashcards: Find Cross-Sections: Triangles & Semicircles

Flashcards: Find Cross-Sections: Triangles & Semicircles

Flashcards: Find Cross-Sections: Triangles & Semicircles

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