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Calculus AB : Calculus AB

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Example Questions

Example Question #2 : Find Cross Sections: Squares & Rectangles

Let be the region bounded by y=x^2-9 and y=0 . Find the volume of the solid whose base is region and whose cross-sections are rectangles perpendicular to the axis with height .

Possible Answers:

278

360

300

265

Correct answer:

360

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: $V=\int_{a}^{b}lw \: dx$ .

Since the region is bounded by y=x^2-9 and y=0 , the base has the following domain: $-3\leq x\leq 3$ .

Putting this all together, we find the following:

$V=\int_{-3}^{3}10(9-x^2) dx =10[9x-\frac{1}{3}x^3] |_{-3}^3=10(27-9+27-9)=360$

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Example Question #712 : Calculus Ab

Let be the region bounded by y=x^2+3 , x=0 and y=12 . Find the volume of the solid whose base is region and whose cross-sections are rectangles perpendicular to the axis with height .

Possible Answers:

108

Correct answer:

108

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 $(V=\int_{a}^{b}A(y) \: dy)$ , we get the following: $V=\int_{a}^{b}lw \: dy$ .

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 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=x2+3 , x=0 and y=12 . The function y=x^2+3 is rewritten in terms of to become $x=\sqrt{y-3}$ , because the final expression should reflect the fact that the cross sections should be written in terms of . Therefore, $w=\sqrt{y-3} -0 =\sqrt{y-3 }$ .

Since the region is bounded by y=x^2+3 and y=12 , the base has the following domain: $3\leq y\leq 12$ .

Putting this all together, we find the following:

$V=\int_{3}^{12}6(\sqrt{y-3}) dy$ $=6\int_{3}^{12}(y-3)^{\frac{1}{2}} dy =6[\frac{2}{3}(y-3)^{\frac{3}{2}}] |_3^{12} =4(27-0)=108$

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Example Question #2 : Find Cross Sections: Squares & Rectangles

Identify the correct expression for the volume of a solid whose base is bounded by y=ln(x)+2 and the axis along $1\leq x\leq 4$ , and whose cross-sections are rectangles perpendicular to the axis with a height three times the width.

Possible Answers:

$3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

$3\int_{1}^{4}ln(x)^2+4 \: dx$

$3\int_{1}^{4}ln(x)+2ln(x) \: dx$

$3\int_{1}^{4}ln(x)^2 \: dx$

Correct answer:

$3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

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 three times the value of the width, or l=3w . The volume expression can now be modified: $V=\int_{a}^{b}3w^2 \: dx$

This just leaves the value of 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=ln(x)+2 and y=0 , along $1\leq x\leq 4$ . Therefore, w=ln(x)+2-0 .

Putting this all together, we find the following:

$V=\int_{1}^{4}3(ln(x)+2)^2 dx=3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

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Example Question #3 : Find Cross Sections: Squares & Rectangles

Identify the correct expression for the volume of a solid whose base is bounded by y=x^5-x^3+4 and y=4 and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

$V=\int_{0}^{4}x^6-2x^8+x^{10} \: dx$

$V=\int_{0}^{4}x^5-x^3-4x \: dx$

$V=\int_{0}^{1}x^3- x^5 \: dx$

Correct answer:

$V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

Explanation:

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

The area of a square is A=s^2 , where is the side length of the square. By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}s^2\: dx$ .

Next, an expression for s^2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s^2 .

s=(4)-(x^5-x^3+4)=x^3-x^5

$s^2=(x^3-x^5)^2=x^6-2x^8+x^{10}$

Since the region is bounded by y=x^5-x^3+4 and y=4 , the base has the following domain: $0\leq x\leq 1$ .

Putting this all together, we find the following:

$V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

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Example Question #1 : Finding Volume Using Integration

Identify the correct expression for the volume of a solid whose base is bounded by y=sin(x) and y=0 along and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$V=\int_{0}^{\pi}sin (x) - \frac{\pi}{2}\: dx$

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

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

$V=\int_{\frac{\pi}{2}}^{\pi}sin^2(y)\: dy$

Correct answer:

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

Explanation:

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

Next, an expression for s^2 must be determined. Because should be the width of the solid’s base, the expression for that length can be used to solve for s^2 .

s=sin(x)-0

s^2=sin^2(x)

Putting this all together, we find the following:

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

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Example Question #1 : Apply The Revolving Disc Method And Washer Method

What is the disc method?

Possible Answers:

Finding the volume of a solid by finding the surface area and adding all surface areas of the solid

Finding the area of a circle

Finding the volume of a sphere

Finding the volume of a solid by taking a cross sectional of it and calculating the volume by adding infinitely many of these cross sections

Correct answer:

Finding the volume of a solid by taking a cross sectional of it and calculating the volume by adding infinitely many of these cross sections

Explanation:

The disc method allows us to find the volume of a solid in a 2 dimensional space by taking the cross section of a solid. When taking the cross section, you are slicing perpendicular to the axis of rotation (usually either x or y axis). Then we sum infinitely many of these cross sections to obtain a value for the volume, sort of like integrating under the curve.

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Example Question #721 : Calculus Ab

What is the formula for the disc method?

Possible Answers:

$V = \pi\int _a^b f(x)^2 dx$

$V = \int_a^b \pi r^2 h dr$

$V = \pi r^2\int_a^b f(x)^2 dx$

$V = \pi r^2 h$

Correct answer:

$V = \pi\int _a^b f(x)^2 dx$

Explanation:

The disc method is based on the volume of a cylinder, $V = \pi r^2 h$ . In the formula for the disc method, is the volume of the solid, is the smallest value of of f(x) , is the largest value of of f(x) , f(x) is the radius of the disc, and is the height of the disc.

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Example Question #1 : Apply The Revolving Disc Method And Washer Method

Find the volume of the solid that is bounded by y = x^2 - 2x + 1 and the x-axis about the x-axis when $0\leq x\leq 2$ .

Possible Answers:

$V = \frac{\pi}{9}$

$V = -2\pi$

$V = \frac{2\pi}{9}$

$V = 2\pi$

Correct answer:

$V = \frac{2\pi}{9}$

Explanation:

First we must find the cross sectional area. To do this, we need the distance from the x-axis to the function (the radius of each disc). If we think about it, this is just the function itself.

$A(x) = \pi f(x)^2$

$A(x) = \pi(x^2-2x+1)^2$

Now we can plug into our disc method formula: $V = \int_a^b A(x) dx$ or $V = \pi\int_a^b f(x)^2dx$ .

$V = \int_a^b A(x) dx$

$V = \pi\int_0^2 (x^2 - 2x + 1)^2 dx$

$V = \pi\int_0^2 ((x-1)^2)^4 dx$

$V = \pi\int_0^2 (x-1)^8 dx$

Use a u-substitution where u = (x-1), du = 1

$V = \pi\int_0^2 u^8 du$

$V = \pi[\frac{u^9}{9}]_0^2$

$V = \pi[\frac{(x-1)^9}{9}]_0^2$

$V = \pi[\frac{(2-1)^9}{9}-\frac{(0-1)^9}{9}]$

$V = \pi[\frac{1}{9} + \frac{1}{9}]$

$V = \frac{2\pi}{9}$

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Example Question #2 : Apply The Revolving Disc Method And Washer Method

Find the volume of the solid using the disc method bounded by y = 5x^3 - 4x and the x-axis when $-1\leq x \leq 4$ about the x-axis. Do not simplify.

Possible Answers:

$V = \pi[\frac{409625}{7} - 7897]$

$V = \pi$

$V = \pi80921$

$V = \pi\frac{409625}{7}$

Correct answer:

$V = \pi[\frac{409625}{7} - 7897]$

Explanation:

First set up the disc method formula then plug in the given function.

$V = \pi\int_a^b f(x)^2 dx$

$V = \pi\int_{-1}^{4} (5x^3 - 4x)^2 dx$

$V = \pi\int_{-1}^{4} 25x^6-40x^4 +14x^2 dx$

$V = \pi[\frac{25x^7}{7} -8x^5 + \frac{14x^3}{3}]_{-1}^{4}$

$V = \pi[(\frac{25(4)^7}{7} - 8(4)^5 + \frac{14(4)^3}{3}) -(\frac{25(-1)^7}{7} - 8(-1)^5 + \frac{14(-1)^3}{3})$

$V = \pi[\frac{409600}{7} - 8193 + \frac{896}{3} +\frac{25}{7} -8 + \frac{14}{3}]$

$V = \pi[\frac{409625}{7} - 8201 + \frac{912}{3}]$

$V = \pi[\frac{409625}{7} - 7897]$

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Example Question #722 : Calculus Ab

True or False: The Washer Method is more accurate than the Disc Method and should be used all the time.

Possible Answers:

True

False

Correct answer:

False

Explanation:

These two methods are used in different situations so their accuracy is not comparable. The Washer Method is like using the disc method when considering the difference between two discs. Their accuracy is similar and you should use the Washer method when you have an inner and outer radius and the disc method when you have only one radius.

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #2 : Find Cross Sections: Squares & Rectangles

Example Question #712 : Calculus Ab

Example Question #2 : Find Cross Sections: Squares & Rectangles

Example Question #3 : Find Cross Sections: Squares & Rectangles

Example Question #1 : Finding Volume Using Integration

Example Question #1 : Apply The Revolving Disc Method And Washer Method

Example Question #721 : Calculus Ab

Example Question #1 : Apply The Revolving Disc Method And Washer Method

Example Question #2 : Apply The Revolving Disc Method And Washer Method

Example Question #722 : Calculus Ab

All Calculus AB Resources

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