Wize AP Calculus (AB) Textbook > Applications of Integration

Exam-Like Practice Problems

Volumes of Revolution (Discs & Washers Method)Previous Section

The acceleration due to gravity of an object is given by a(t)=−9.8a\left(t\right)=-9.8a(t)=−9.8. If the initial velocity and displacement are given by v(0)=5v\left(0\right)=5v(0)=5 and d(0)=2d\left(0\right)=2d(0)=2, find the function that represents the displacement at time t.

d(t)=−4.9t2+5t+2d\left(t\right)=-4.9t^2+5t+2d(t)=−4.9t2+5t+2

d(t)=−4.9t2−5t+2d\left(t\right)=-4.9t^2-5t+2d(t)=−4.9t2−5t+2

d(t)=−9.8t+2d\left(t\right)=-9.8t+2d(t)=−9.8t+2

d(t)=−9.8t2+5t+2d\left(t\right)=-9.8t^2+5t+2d(t)=−9.8t2+5t+2

d(t)=−9.8t2−5t+2d\left(t\right)=-9.8t^2-5t+2d(t)=−9.8t2−5t+2

I don't know

If the displacement of a particle is given by s(t)=∫0tx2sin⁡xdxs\left(t\right)=\int_0^t\frac{x^2}{\sin x}dxs(t)=∫0t​sinxx2​dx, then the acceleration at at t=π2t=\frac{\pi}{2}t=2π​ is

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find the average value of the function g(x)=ax+xag\left(x\right)=a^x+x^ag(x)=ax+xa  over the interval [2,3]\left[2,3\right][2,3]  , where a ≠−1a\ \ne-1a =−1  is a constant.

I have answered this question

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find the average value of y=x2y=x^2y=x2 over [1,3][1, 3][1,3] .

I have answered this question

Find the average value of f(x)=sin⁡xf(x)=\sin xf(x)=sinx on [0,π/2][0,\pi/2][0,π/2].

Answer

I don't know

Find the area of the region bounded between x2−x+1x^2-x+1x2−x+1 and the line y=1y=1y=1.

Answer

I don't know

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find integral that represents the area enclosed by the curves y=x3y=x^3y=x3 and y=xy=xy=x.

I have answered this question

 Find the area bounded by the curves  y=−x  and  y= x2 −2.y=-x\ \text{ and }\ y=\ x^2\ -2.y=−x  and  y= x2 −2.

Answer

I don't know

Find the area bounded by the curves y=cos⁡xy=\cos xy=cosx and y=sin⁡xy=\sin xy=sinx from 0 to π/2\pi/2π/2. Sketch the area.

Answer

I don't know

Find the volume of the solid that is produced by revolving the region bounded between x=1+y2x=1+y^2x=1+y2, x=0x=0x=0, y=1y=1y=1, and y=2y=2y=2 about the xxx-axis.

Answer

I don't know

Find the volume of the solid that is produced by revolving y=xsin⁡xcos⁡xy=\sqrt{x\sin x\cos x}y=xsinxcosx​ about the xxx-axis, between 0 and π/2\pi/2π/2.

Answer

I don't know

Find the volume of the solid that is produced by revolving the region bounded between y=x2/4y=x^2/4y=x2/4, y=0y=0y=0, and x=2x=2x=2 about the yyy-axis. Use the method of cylindrical shells.

Answer

I don't know

Write the integrals representing the volume of the solid that is produced by revolving the region bounded between y=tan⁡3xy=\tan^3xy=tan3x, y=1y=1y=1, and x=0x=0x=0 about the line y=2y=2y=2 using both the washer method and the method of cylindrical shells. Do not compute the integrals.

Washer: V=π∫0π/4(2−tan⁡3x)2−(2−1)2dxV = \pi\displaystyle\int^{\pi/4}_0(2 − \tan^3x)^2 − (2 − 1)^2 dxV=π∫0π/4​(2−tan3x)2−(2−1)2dx

Shells: V=2π∫01(2−y)arctan⁡(y3)dyV = 2\pi\displaystyle\int^1_0(2 − y)\arctan(\sqrt[3]{y}) dyV=2π∫01​(2−y)arctan(3y​)dy

Washer: V=π∫0π/4(2+tan⁡3x)2−(2−1)2dxV=\pi\displaystyle\int_0^{\pi/4}(2+\tan^3x)^2−(2−1)^2dxV=π∫0π/4​(2+tan3x)2−(2−1)2dx

Shells: V=2π∫01(2−y)arctan⁡(y3)dyV=2\pi\displaystyle\int_0^1(2−y)\arctan(\sqrt[3]{y})dyV=2π∫01​(2−y)arctan(3y​)dy

Washer: V=π∫0π/4(2+tan⁡3x)2−(2−1)2dxV=\pi\displaystyle\int_0^{\pi/4}(2+\tan^3x)^2−(2−1)^2dxV=π∫0π/4​(2+tan3x)2−(2−1)2dx

Shells: V=2π∫01(2+y)arctan⁡(y3)dyV=2\pi\displaystyle\int_0^1(2+y)\arctan(\sqrt[3]{y})dyV=2π∫01​(2+y)arctan(3y​)dy

Washer: V=π∫0π/4(2−tan⁡3x)2−(2−1)2dxV=\pi\displaystyle\int_0^{\pi/4}(2−\tan^3x)^2−(2−1)^2dxV=π∫0π/4​(2−tan3x)2−(2−1)2dx

Shells: V=2π∫01(2+y)arctan⁡(y3)dyV=2\pi\displaystyle\int_0^1(2+y)\arctan(\sqrt[3]{y})dyV=2π∫01​(2+y)arctan(3y​)dy

None of the above

I don't know

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find the arc length function for the curve y=2x3/2y=2x^{3/2}y=2x3/2 with starting point P0=(0,0)P_0=\left(0,0\right)P0​=(0,0).

I have answered this question

Practice Question
Find the arc length function for the curve y=2x32y=2x^{\frac{3}{2}}y=2x23​ with starting point P0(0,3)P_0\left(0,3\right)P0​(0,3).

Arc length

s(x)=

I don't know

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find the arc length of y=exy=e^xy=ex on [0,1][0,1][0,1].

I have answered this question

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Q:\textbf{Q:}Q: Consider the curve  y=f(x)=x33+14xy=f(x)=\dfrac{x^3}{3}+\dfrac{1}{4x}y=f(x)=3x3​+4x1​. Find the arc length of this curve over the interval [1,3][1, 3][1,3]. Find a function giving the arc length of the curve over the interval [1,x].[1, x].[1,x].

I have answered this question

Set up the appropriate integral for determining the surface area of revolution for the following curve: y=x2\bcb{y = x^2}y=x2, (0≤x≤2\bcb{0 \leq x \leq 2}0≤x≤2), about the y\bcb{y}y-axis.

2π∫02x1+4x2  ⁣dx.\displaystyle { 2\pi\int_0^{2} x\sqrt{1+4x^2}  \de{x}}.2π∫02​x1+4x2​ dx.

2π∫02x1+2x2  ⁣dx.\displaystyle { 2\pi\int_0^{2} x\sqrt{1+2x^2}  \de{x}}.2π∫02​x1+2x2​ dx.

2π∫02x1+2x  ⁣dx.\displaystyle { 2\pi\int_0^{2} x\sqrt{1+2x}  \de{x}}.2π∫02​x1+2x​ dx.

None of the other options.

I don't know

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find the area of the surface obtained by revolving the curve 
y=9−x2y = \sqrt{9 -x^2}y=9−x2​
from x=−3x = -3x=−3 to x=3x = 3x=3 about the xxx - axis.

I have answered this question

Wize AP Calculus (AB) Textbook > Applications of Integration

Exam-Like Practice Problems

Popular Courses

Practice Question

Set up the appropriate integral for determining the surface area of revolution for the following curve: $\bcb{y = x^2}$ , ( $\bcb{0 \leq x \leq 2}$ ), about the $\bcb{y}$ -axis.