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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

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Example Questions

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Find the expression for the volume of the solid whose cross-sections are semicircles perpendicular to the axis and whose base is bounded by y=x^2-1 and y=x+1 .

Possible Answers:

$V=\frac{1}{8} \int_{0}^{3}2+x-x^2\: dx$

$V=\frac{1}{8}\pi\int_{-1}^{2}(2+x-x^2)^2 dx$

$V=\int_{-1}^{2}(2+x-x^2)^2 \: dx$

$V= \pi\int_{-1}^{2} x^2 - 1 \: dx$

Correct answer:

$V=\frac{1}{8}\pi\int_{-1}^{2}(2+x-x^2)^2 dx$

Explanation:

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

The area of a semicircle is $A=\frac{1}{2} \pi r^2$ . By applying this formula to our general volume formula $(V= \int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}\frac{1}{2} \pi r^2\: dx$ .

Since the region bounded by y=x^2-1 and y=x+1 is the base of the solid, the intersection points of these functions will create the bounds for the volume expression. These points are (-1,0) and (2,3) . Since the expression is in terms of , the coordinates can be referenced for the bounds.

Next, an expression for must be determined. Since the radius is half the diameter of the semicircle, and the diameter of the semicircle is the length stretching between the functions y=x^2-1 and y=x+1 , the expression of the radius is the following: $r=\frac{1}{2}[(x+1)-(x^2-1)]$ . Simplified, this reads $r=\frac{1}{2}(2+x-x^2)$ .

Putting this all together, we find the following:

$V=\int_{-1}^{2}\frac{1}{2} \pi (\frac{1}{2}(2+x-x^2))^2 \: dx =\frac{1}{8}\pi \int_{-1}^{2}(2+x-x^2)^2\: dx$

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Example Question #2 : Find Cross Sections: Triangles & Semicircles

Find the volume of the solid whose cross-sections are semicircles and whose base is bounded by the circle x^2+y^2=9 .

Possible Answers:

$36 \pi$

$18 \pi$

Correct answer:

$18 \pi$

Explanation:

The base is defined by the following formula: x^2+y^2=9 . Therefore, the radius of the base is . The radius defines the bounds as being [-3,3]

The correct formula for the area of a semicircle is as follows:

$A=\frac{1}{2} \pi r^2$ , with r being the radius of the semicircle.

By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x)\: dx)$ , we get the following: $V=\int_{a}^{b}\frac{1}{2} \pi r^2 \: dx$ .

Next, an expression for must be determined. The radius is half the diameter of the semicircle cross-section. The value of is equivalent to the half the height of the base, or $y=\sqrt{9-x^2}$ . Therefore, $r=\sqrt{9-x^2}$ .

Putting this all together, we find the following:

$V=\int_{-3}^{3}\frac{1}{2} \pi (\sqrt{9-x^2})^2\: dx$ $=\frac{1}{2} \pi \int_{-3}^{3}9-x^2\: dx =\frac{1}{2} \pi [9x-\frac{1}{3}x^3] |_{-3}^3=\frac{1}{2} \pi (27-9+27-9) =18 \pi$

*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.

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Example Question #1 : Find Cross Sections: Triangles & Semicircles

Identify the correct expression for the volume of the solid whose cross-sections are semicircles perpendicular to the axis and whose base is bounded by y=x^2 and y=4 .

Possible Answers:

$V= \pi \int_{-2}^{2}16 + x^4 - 8x^6\: dx$

$V=\frac{1}{8} \int_{-2}^{2} 8x^2+x^4\: dx$

$V=\frac{1}{8} \pi \int_{-2}^{2}16-8x^2+x^4\: dx$

$V= \pi \int_{-2}^{4}16-8x^2+x^4\: dx$

Correct answer:

$V=\frac{1}{8} \pi \int_{-2}^{2}16-8x^2+x^4\: dx$

Explanation:

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

The area of a semicircle is $A=\frac{1}{2} \pi r^2$ . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}\frac{1}{2} \pi r^2 \: dx$ .

Since the region bounded by y=x^2 and y=4 is the base of the solid, the intersection points of these functions will create the bounds for the volume expression. These points are (-2,4) and (2,4) . Since the expression is in terms of , the coordinates can be referenced for the bounds.

Next, an expression for must be determined. Since the radius is half the diameter of the semicircle, and the diameter of the semicircle is the length stretching between the functions y=x^2 and y=4 , the expression of the radius is the following: $r=\frac{1}{2}(4-x^2)$ .

Putting this all together, we find the following:

$V=\int_{-2}^{2}\frac{1}{2} \pi (\frac{1}{2}(4-x))^2 \: dx =\frac{1}{8} \pi \int_{-2}^{2}(4-x^2)^2 \: dx =\frac{1}{8} \pi \int_{-2}^{2}16-8x^2+x^4\: dx$

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Example Question #1 : Find Cross Sections: Triangles & Semicircles

Identify the correct expression for the volume of the solid whose cross-sections are semicircles perpendicular to the axis and whose base is bounded by y=x^2+3 and y=5-x .

Possible Answers:

$V= \int_{-1}^{1}(2-x^2)^2 \: dx$

$V=\frac{1}{8} \pi \int_{-1}^{1}(2-x-x^2)^2 \: dx$

$V=\frac{1}{8} \pi \int_{-2}^{1}(2-x-x^2)^2 \: dx$

$V=\pi \int_{-2}^{1}(2-x^2)^2 \: dx$

Correct answer:

$V=\frac{1}{8} \pi \int_{-2}^{1}(2-x-x^2)^2 \: dx$

Explanation:

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

The area of a semicircle is $A=\frac{1}{2} \pi r^2$ . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V= \int_{a}^{b} \frac{1}{2} \pi r^2 \: dx$ .

Since the region bounded by y=x^2+3 and y=5-x is the base of the solid, the intersection points of these functions will create the bounds for the volume expression. These points are (-2,7) and (1,4) . Since the expression is in terms of , the coordinates can be referenced for the bounds.

Next, an expression for must be determined. Since the radius is half the diameter of the semicircle, and the diameter of the semicircle is the length stretching between the functions y=x^2+3 and y=5-x , the expression of the radius is the following: $r=\frac{1}{2}[(5-x)-(x^2+3)]$ . This can be simplified: $r=\frac{1}{2}(2-x-x^2)$

Putting this all together, we find the following:

$V=\int_{-2}^{1}\frac{1}{2} \pi (\frac{1}{2}(2-x-x^2))^2 \: dx =\frac{1}{8} \pi \int_{-2}^{1}(2-x-x^2)^2\: dx$

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Example Question #741 : Calculus Ab

Identify the correct expression for the volume of the solid whose cross-sections are semicircles parallel to the y axis and whose base is bounded by y=ln(x) , x=2 and x=6 .

Possible Answers:

$V=\frac{1}{8} \pi \int_{2}^{6}ln(x)^2 \: dx$

$V= \pi \int_{2}^{6} 2 ln(x) \: dx$

$V=\frac{1}{8} \pi \int_{2}^{6} e^{2y} \: dy$

$V=\pi \int_{2}^{6} ln(x) \: dx$

Correct answer:

$V=\frac{1}{8} \pi \int_{2}^{6}ln(x)^2 \: dx$

Explanation:

The cross-sections are parallel to the axis; this is another way of saying the cross-sections are perpendicular to the axis. Therefore, the volume expression will be in terms of .

The area of a semicircle is $A=\frac{1}{2} \pi r^2$ . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}\frac{1}{2} \pi r^2 \: dx$ .

Since the region is bounded by y=ln(x) , x=2 , and x=6 , the base is the area between the axis and y=ln(x) on the interval $2\leq x \geq 6$ . Since the expression is in terms of , the interval $2\leq x \geq 6$ will define the bounds.

Next, an expression for must be determined. Since the radius is half the diameter of the semicircle, and the diameter of the semicircle is the length stretching between y=ln(x) and the axis, the expression of the radius is the following: $r=\frac{1}{2}ln(x)$ .

Putting this all together, we find the following:

$V=\int_{2}^{6}\frac{1}{2} \pi (\frac{1}{2}(ln(x))^2 dx =\frac{1}{8} \pi \int_{2}^{6}ln(x)^2 \: dx$

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Example Question #1 : 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 .

Possible Answers:

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

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

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

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

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:

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

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Example Question #1 : 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.

Possible Answers:

$\frac{56}{3}$

$\frac{64}{3}$

$\frac{64}{5}$

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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Example Question #1 : Find Cross Sections: Triangles & Semicircles

Identify the correct expression for the volume of the solid whose cross-sections are equilateral triangles perpendicular to the axis and whose base is bounded by y=x^2 and y=x .

Possible Answers:

$V=\int_{0}^{1}\frac{\sqrt{3}}{4}(\sqrt{x}-x)^2 dx$

$V=\int_{0}^{1}\frac{\sqrt{3}}{4}(\sqrt{y}-y) \: dy$

$V=\int_{0}^{1}\sqrt{x}-x \: dx$

$V=\int_{0}^{1}\frac{\sqrt{3}}{4}(\sqrt{y}-y)^2 dy$

Correct answer:

$V=\int_{0}^{1}\frac{\sqrt{3}}{4}(\sqrt{y}-y)^2 dy$

Explanation:

First, the cross sections being perpendicular to the axis indicates the expression should be in terms of .

The area of an equilateral triangle is $A=\frac{\sqrt{3}}{4}s^2$ , with being the side length of the triangle. By applying this formula to our general volume formula ( $V=\int_{a}^{b}A(y) dy$ ), we get the following: $V=\int_{a}^{b}\frac{\sqrt{3}}{4}s^2 dy$ .

The intersection points of the functions y=x^2 and y=x are (0,0) and (1,1) . The coordinates of these points will define the bounds for the integral, since our expression is in terms of .

The base is bounded by y=x^2 and y=x . Rewriting these functions in terms of , the following equations are obtained: $x=\sqrt{y}$ and x=y . Since $x=\sqrt{y}$ is farther from the axis, the correct expression for the side length is $s=\sqrt{y}-y$ .

Putting it all together, the following is obtained:

$V=\int_{0}^{1}\frac{\sqrt{3}}{4}(\sqrt{y}-y)^2 \: dy$

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Example Question #1 : Find Cross Sections: Triangles & Semicircles

Identify the correct expression for the volume of the solid whose base is bounded by y=x^3 , y=8 , and x=0 , and whose cross-sections are right isosceles triangles, perpendicular to the axis, with one leg on the base of the solid.

Possible Answers:

$V=\int_{0}^{2}8-x^3 dx$

$V=\int_{0}^{2}\frac{1}{2}(x^6) dx$

$V=\int_{0}^{8}\frac{1}{2}(8-x^3) dx$

$V=\int_{0}^{2}\frac{1}{2}(8-x^3)^2 dx$

Correct answer:

$V=\int_{0}^{2}\frac{1}{2}(8-x^3)^2 dx$

Explanation:

First, the cross sections being perpendicular to the axis indicates the expression should be in terms of . 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 intersection points of the functions defining the region are (2,8) and (0,0) . The coordinates of these points will define the bounds for the integral, since our expression is in terms of .

The base is bounded by y=x^3 , x=0 and y=8 . Since the cross-sections are perpendicular to the axis, the leg of the triangle cross-sections are defined by: b=8-x^3 .

Putting it all together, the following is obtained:

$V=\int_{0}^{2}\frac{1}{2}(8-x^3)^2\: dx$

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Example Question #2 : 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 .

Possible Answers:

$36\sqrt{3}$

$6\sqrt{2}$

$6\sqrt{3}$

Correct answer:

$36\sqrt{3}$

Explanation:

Because the disk is of radius , the base is defined by the following formula: x^2+y^2=9 .

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

$A=\frac{\sqrt{3}}{4}s^2$ , with s 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 defines the bounds as being [-3,3] . Next, 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 [-3,3] . The length of one side of the equilateral triangle, therefore, is $s=2\sqrt{9-x^2}$ .

Putting it all together, the following is obtained:

$V=\int_{-3}^{3}\frac{\sqrt{3}}{4}(2\sqrt{9-x^2})^2 dx=\int_{-3}^{3}\sqrt{3 }(9-x^2) dx$ $=[\sqrt{3}(9x-\frac{1}{3}x^3)] |_{-3}^3= \sqrt{3}(27-9+27-9)=36\sqrt{3}$

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #2 : Find Cross Sections: Triangles & Semicircles

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #741 : Calculus Ab

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #1 : Find Cross Sections: Triangles & Semicircles

Example Question #2 : Find Cross Sections: Triangles & Semicircles

All Calculus AB Resources

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