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Calculus AB : 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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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

All Calculus AB Resources

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