# 1.1E: Exercises

- Page ID
- 18539

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## Determining Volumes by Slicing

### Exercise \(\PageIndex{1}\)

Derive the formula for the volume of a sphere using the slicing method.

**Answer**-
As in most problems we should choose to graph the function first. We know the formula for a circle, \(x^2+y^2=r^2\) and solving for y we get \(y=\pm\sqrt{R^2-x^2}\). When graphed they appear as:

With the blue line being the positive values and the green the negatives. Now we must find what an area of a slice will be. As we are trying to find the volume of a sphere the area of a slice must be a circle, with radius \(y=r=\sqrt{R^2-x^2}\).

Setting up the area of a slice we get \[A(x)=\pi r^2=\pi (\sqrt{R^2-x^2})^2=\pi(R^2-x^2)\]

Integrating over the interval of \([-R,R]\) we get: \[V=\pi \int_{-R}^{R}(R^2-x^2)dx=\pi \Big[R^2x-\dfrac{1}{3}x^3\Big]_{-R}^{R}=\dfrac{4}{3}\pi R^3\]

Exercise \(\PageIndex{2}\)

Derive the formula for the volume of a cone.

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{3}\)

Use the slicing method to derive the formula for the volume of a tetrahedron with side length a.

**Answer**-
TBD.

### Exercise \(\PageIndex{4}\)

Use the disk method to derive the formula for the volume of a trapezoidal cylinder.

**Answer**-
TBD

### Exercise \(\PageIndex{5}\)

Explain when you would use the disk method versus the washer method. When are they interchangeable?

**Answer**-
TBD

**For exercises 6 - 10, draw a typical slice and find the volume using the slicing method for the given volume.**

### Exercise \(\PageIndex{6}\)

A pyramid with height 6 units and square base of side 2 units, as pictured here.

**Answer**-
Here the cross-sections are squares taken perpendicular to the \(y\)-axis.

We use the vertical cross-section of the pyramid through its center to obtain an equation relating \(x\) and \(y\).

Here this would be the equation, \( y = 6 - 6x \). Since we need the dimensions of the square at each \(y\)-level, we solve this equation for \(x\) to get, \(x = 1 - \tfrac{y}{6}\).

This is half the distance across the square cross-section at the \(y\)-level, so the side length of the square cross-section is, \(s = 2\left(1 - \tfrac{y}{6}\right).\)

Thus, we have the area of a cross-section is,

\(A(y) = \left[2\left(1 - \tfrac{y}{6}\right)\right]^2 = 4\left(1 - \tfrac{y}{6}\right)^2.\)

\(\begin{align*} \text{Then},\quad V &= \int_0^6 4\left(1 - \tfrac{y}{6}\right)^2 \, dy \\[5pt]

&= -24 \int_1^0 u^2 \, du, \quad \text{where} \, u = 1 - \tfrac{y}{6}, \, \text{so} \, du = -\tfrac{1}{6}\,dy, \quad \implies \quad -6\,du = dy \\[5pt]

&= 24 \int_0^1 u^2 \, du = 24\dfrac{u^3}{3}\bigg|_0^1 \\[5pt]

&= 8u^3\bigg|_0^1 \\[5pt]

&= 8\left( 1^3 - 0^3 \right) \quad= \quad 8\, \text{units}^3 \end{align*}\)

### Exercise \(\PageIndex{7}\)

A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.

**Answer**-
TBD

### Exercise \(\PageIndex{8}\)

A tetrahedron with a base side of 4 units,as seen here.

**Answer**-
\(V = \frac{32}{3\sqrt{2}} = \frac{16\sqrt{2}}{3}\) units

^{3}

### Exercise \(\PageIndex{9}\)

A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.

**Answer**-
TBD

### Exercise \(\PageIndex{10}\)

A cone of radius \( r\) and height \( h\) has a smaller cone of radius \( r/2\) and height \( h/2\) removed from the top, as seen here. The resulting solid is called a frustum.

**Answer**-
\(V = \frac{7\pi}{12} hr^2\) units

^{3}

**For exercises 11 - 16, draw an outline of the solid and find the volume using the slicing method.**

### Exercise \(\PageIndex{11}\)

The base is a circle of radius \( a\). The slices perpendicular to the base are squares.

**Answer**-
TBD

### Exercise \(\PageIndex{12}\)

The base is a triangle with vertices \( (0,0),(1,0),\) and \( (0,1)\). Slices perpendicular to the \(xy\)-plane are semicircles.

**Answer**-
\(\displaystyle V = \int_0^1 \frac{\pi(1-x)^2}{8}\, dx \quad = \quad \frac{π}{24}\) units

^{3}

### Exercise \(\PageIndex{13}\)

The base is the region under the parabola \( y=1−x^2\) in the first quadrant. Slices perpendicular to the \(xy\)-plane are squares.

**Answer**-
TBD

### Exercise \(\PageIndex{14}\)

The base is the region under the parabola \( y=1−x^2\) and above the **\(x\)-axis**. Slices perpendicular to the** \(y\)-axis** are squares.

**Answer**-
\(\displaystyle V = \int_0^1 4(1 - y)\,dy \quad = \quad 2\) units

^{3}

### Exercise \(\PageIndex{15}\)

The base is the region enclosed by \( y=x^2)\) and \( y=9.\) Slices perpendicular to the \(x\)-axis are right isosceles triangles.

**Answer**-
TBD

### Exercise \(\PageIndex{16}\)

The base is the area between \( y=x\) and \( y=x^2\). Slices perpendicular to the \(x\)-axis are semicircles.

**Answer**-
\(\displaystyle V = \int_0^1 \frac{\pi}{8}\left( x - x^2 \right)^2 \, dx \quad=\quad \frac{π}{240}\) units

^{3}

## Disk and Washer Method

**For exercises 17 - 24, draw the region bounded by the curves. Then, use the disk or washer method to find the volume when the region is rotated around the \(x\)-axis.**

### Exercise \(\PageIndex{17}\)

\( x+y=8,\quad x=0\), and \( y=0\)

**Answer**-
TBD

### Exercise \(\PageIndex{18}\)

\( y=2x^2,\quad x=0,\quad x=4,\) and \( y=0\)

**Answer**-
\(\displaystyle V = \int_0^4 4\pi x^4\, dx \quad=\quad \frac{4096π}{5}\) units

^{3}

### Exercise \(\PageIndex{19}\)

\( y=e^x+1,\quad x=0,\quad x=1,\) and \( y=0\)

**Answer**-
TBD

### Exercise \(\PageIndex{20}\)

\( y=x^4,\quad x=0\), and \( y=1\)

**Answer**-
\(\displaystyle V = \int_0^1 \pi\left( 1^2 - \left( x^4\right)^2\right)\, dx = \int_0^1 \pi\left( 1 - x^8\right)\, dx \quad = \quad \frac{8π}{9}\) units

^{3}

### Exercise \(\PageIndex{21}\)

\( y=\sqrt{x},\quad x=0,\quad x=4,\) and \( y=0\)

**Answer**-
TBD

### Exercise \(\PageIndex{22}\)

\( y=\sin x,\quad y=\cos x,\) and \( x=0\)

**Answer**-
\(\displaystyle V = \int_0^{\pi/4} \pi \left( \cos^2 x - \sin^2 x\right) \, dx = \int_0^{\pi/4} \pi \cos 2x \, dx \quad=\quad \frac{π}{2}\) units

^{3}

### Exercise \(\PageIndex{23}\)

\( y=\dfrac{1}{x},\quad x=2\), and \( y=3\)

**Answer**-
TBD

### Exercise \(\PageIndex{24}\)

\( x^2−y^2=9\) and \( x+y=9,\quad y=0\) and \( x=0\)

**Answer**-
\(V = 207π\) units

^{3}

**For exercises 25 - 32, draw the region bounded by the curves. Then, find the volume when the region is rotated around the \(y\)-axis.**

### Exercise \(\PageIndex{25}\)

\( y=4−\dfrac{1}{2}x,\quad x=0,\) and \( y=0\)

**Answer**-
TBD

### Exercise \(\PageIndex{26}\)

\( y=2x^3,\quad x=0,\quad x=1,\) and \( y=0\)

**Answer**-
\(V = \frac{4π}{5}\) units

^{3}

### Exercise \(\PageIndex{27}\)

\( y=3x^2,\quad x=0,\) and \( y=3\)

**Answer**-
TBD

### Exercise \(\PageIndex{28}\)

\( y=\sqrt{4−x^2},\quad y=0,\) and \( x=0\)

**Answer**-
\(V = \frac{16π}{3}\) units

^{3}

### Exercise \(\PageIndex{29}\)

\( y=\dfrac{1}{\sqrt{x+1}},\quad x=0\), and \( x=3\)

**Answer**-
TBD

### Exercise \(\PageIndex{30}\)

\( x=\sec(y)\) and \( y=\dfrac{π}{4},\quad y=0\) and \( x=0\)

**Answer**-
\(V = π\) units

^{3}

### Exercise \(\PageIndex{31}\)

\( y=\dfrac{1}{x+1},\quad x=0\), and \( x=2\)

**Answer**-
TBD

### Exercise \(\PageIndex{32}\)

\( y=4−x,\quad y=x,\) and \( x=0\)

**Answer**-
\(V = \frac{16π}{3}\) units

^{3}

**For exercises 33 - 40, draw the region bounded by the curves. Then, find the volume when the region is rotated around the \(x\)-axis.**

### Exercise \(\PageIndex{33}\)

\( y=x+2,\quad y=x+6,\quad x=0\), and \( x=5\)

**Answer**-
TBD

### Exercise \(\PageIndex{34}\)

\( y=x^2\) and \( y=x+2\)

**Answer**-
\(V = \frac{72π}{5}\) units

^{3}

### Exercise \(\PageIndex{35}\)

\( x^2=y^3\) and \( x^3=y^2\)

**Answer**-
TBD

### Exercise \(\PageIndex{36}\)

\( y=4−x^2\) and \( y=2−x\)

**Answer**-
\(V = \frac{108π}{5}\) units

^{3}

### Exercise \(\PageIndex{37}\)

\( y=\cos x,\quad y=e^{−x},\quad x=0\), and \( x=1.2927\)

**Answer**-
TBD

### Exercise \(\PageIndex{38}\)

\( y=\sqrt{x}\) and \( y=x^2\)

**Answer**-
\(V = \frac{3π}{10}\) units

^{3}

### Exercise \(\PageIndex{39}\)

\( y=\sin x,\quad y=5\sin x,\quad x=0\) and \( x=π\)

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{40}\)

\( y=\sqrt{1+x^2}\) and \( y=\sqrt{4−x^2}\)

**Answer**-
\(V = 2\sqrt{6}π\) units

^{3}

**For exercises 41 - 45, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the \(y\)-axis.**

### Exercise \(\PageIndex{41}\)

\( y=\sqrt{x},\quad x=4\), and \( y=0\)

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{42}\)

\( y=x+2,\quad y=2x−1\), and \( x=0\)

**Answer**-
\(V = 9π\) units

^{3}

### Exercise \(\PageIndex{43}\)

\( y=\dfrac{3}{x}\) and \( y=x^3\)

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{44}\)

\( x=e^{2y},\quad x=y^2,\quad y=0\), and \( y=\ln(2)\)

**Answer**-
\(V = \dfrac{π}{20}(75−4\ln^5(2))\) units

^{3}

### Exercise \(\PageIndex{45}\)

\( x=\sqrt{9−y^2},\quad x=e^{−y},\quad y=0\), and \( y=3\)

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{46}\)

Yogurt containers can be shaped like frustums. Rotate the line \( y=\left(\frac{1}{m}\right)x\) around the \(y\)-axis to find the volume between \( y=a\) and \( y=b\).

**Answer**-
\(V = \dfrac{m^2π}{3}(b^3−a^3)\) units

^{3}

### Exercise \(\PageIndex{47}\)

Rotate the ellipse \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \) around the \(x\)-axis to approximate the volume of a football, as seen here.

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{48}\)

Rotate the ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) around the \(y\)-axis to approximate the volume of a football.

**Answer**-
\(V = \frac{4a^2bπ}{3}\) units

^{3}

### Exercise \(\PageIndex{49}\)

A better approximation of the volume of a football is given by the solid that comes from rotating \( y=\sin x) around the \(x\)-axis from \( x=0\) to \( x=π\). What is the volume of this football approximation, as seen here?

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{50}\)

What is the volume of the Bundt cake that comes from rotating \( y=\sin x\) around the \(y\)-axis from \( x=0\) to \( x=π\)?

**Answer**-
\(V = 2π^2\) units

^{3}

**For exercises 51 - 56, find the volume of the solid described.**

### Exercise \(\PageIndex{51}\)

The base is the region between \( y=x\) and \( y=x^2\). Slices perpendicular to the \(x\)-axis are semicircles.

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{52}\)

The base is the region enclosed by the generic ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\) Slices perpendicular to the \(x\)-axis are semicircles.

**Answer**-
\(V = \frac{2ab^2π}{3}\) units

^{3}

### Exercise \(\PageIndex{53}\)

Bore a hole of radius a down the axis of \(a\) right cone and through the base of radius \(b\), as seen here.

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{54}\)

Find the volume common to two spheres of radius \(r\) with centers that are \(2h\) apart, as shown here.

**Answer**-
\(V = \frac{π}{12}(r+h)^2(6r−h)\) units

^{3}

### Exercise \(\PageIndex{55}\)

Find the volume of a spherical cap of height \(h\) and radius \(r\) where \(h<r\), as seen here.

**Answer**-
Add texts here. Do not delete this text first.

### Exercise \(\PageIndex{56}\)

Find the volume of a sphere of radius \(R\) with a cap of height \(h\) removed from the top, as seen here.

**Answer**-
\(V = \dfrac{π}{3}(h+R)(h−2R)^2\) units

^{3}