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Calculus 1 : Regions

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Example Question #41 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 2 and a major axes(the horizontal length) of 10, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 3 units to the left of the origin and the second cut is 2 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{224}{75}\pi$

$\frac{248}{75}\pi$

$\frac{88}{75}\pi$

$\frac{203}{75}\pi$

$\frac{496}{75}\pi$

Correct answer:

$\frac{224}{75}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=2 . and the length of the spheroid 2a=10 , or a=5 :

$V=\pi (2)^2(x-\frac{x^3}{3(5)^2})|_{-3}^{-2}$

$V=\pi (2)^2((-2)-\frac{(-2)^3}{3(5)^2})-\pi (2)^2((-3)-\frac{(-3)^3}{3(5)^2})$

$V=\frac{224}{75}\pi$

Report an Error

Example Question #42 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 1 and a major axes(the horizontal length) of 12, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 5 units to the left of the origin and the second cut is 4 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{47}{54}\pi$

$\frac{47}{108}\pi$

$\frac{235}{54}\pi$

$\frac{47}{36}\pi$

$\frac{235}{108}\pi$

Correct answer:

$\frac{47}{108}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=1 . and the length of the spheroid 2a=12 , or a=6 :

$V=\pi (x-\frac{x^3}{3(6)^2})|_{-5}^{-4}$

$V=\pi (-4-\frac{(-4)^3}{3(6)^2})-\pi (-5-\frac{(-5)^3}{3(6)^2})$

$V=\frac{47}{108}\pi$

Report an Error

Example Question #43 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 2 and a major axes(the horizontal length) of 6, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 2 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{368}{9}\pi$

$\frac{368}{27}\pi$

$\frac{194}{9}\pi$

$\frac{184}{27}\pi$

$\frac{413}{27}\pi$

Correct answer:

$\frac{368}{27}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

$V=\pi (2)^2(x-\frac{x^3}{3(3)^2})|_{-2}^{2}$

$V=\pi (2)^2(2-\frac{2^3}{3(3)^2})-\pi (2)^2(-2-\frac{(-2)^3}{3(3)^2})$

$V=\frac{368}{27}\pi$

Report an Error

Example Question #44 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 1 and a major axes(the horizontal length) of 50, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 14 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{18248}{1875}\pi$

$\frac{7136}{1875}\pi$

$\frac{27248}{1875}\pi$

$\frac{2251}{1875}\pi$

$\frac{27394}{1875}\pi$

Correct answer:

$\frac{27248}{1875}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

$V=\pi (x-\frac{x^3}{3(25)^2})|_{-2}^{14}$

$V=\pi (14-\frac{14^3}{3(25)^2})-\pi (-2-\frac{(-2)^3}{3(25)^2})$

$V=\frac{27248}{1875}\pi$

Report an Error

Example Question #45 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 8 and a major axes(the horizontal length) of 20, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is at the origin and the second cut is 5 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{440}{3}\pi$

$\frac{880}{3}\pi$

$\frac{1760}{3}\pi$

$\frac{110}{3}\pi$

$\frac{220}{3}\pi$

Correct answer:

$\frac{880}{3}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=8 . and the length of the spheroid 2a=20 , or a=10 :

$V=\pi (8)^2(x-\frac{x^3}{3(10)^2})|_{0}^{5}$

$V=\pi (8)^2(5-\frac{5^3}{3(10)^2})-\pi (8)^2(0-\frac{0^3}{3(10)^2})$

$V=\frac{880}{3}\pi$

Report an Error

Example Question #46 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 2 and a major axes(the horizontal length) of 5, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 1 unit to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{208}{75}\pi$

$\frac{104}{75}\pi$

$\frac{416}{75}\pi$

$\frac{188}{75}\pi$

$\frac{181}{75}\pi$

Correct answer:

$\frac{188}{75}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

$V=\pi (2)^2(x-\frac{x^3}{3(\frac{5}{2})^2})|_{-2}^{-1}$

$V=\pi (2)^2(-1-\frac{(-1)^3}{3(\frac{5}{2})^2})-\pi (2)^2(-2-\frac{(-2)^3}{3(\frac{5}{2})^2})$

$V=\frac{188}{75}\pi$

Report an Error

Example Question #47 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 1 and a major axes (the horizontal length) of 3, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 1 unit to the left of the origin and the second cut is at the origin, what is the volume of the segment?

Possible Answers:

$\frac{19}{27}\pi$

$\frac{28}{27}\pi$

$\frac{23}{27}\pi$

$\frac{17}{27}\pi$

$\frac{11}{27}\pi$

Correct answer:

$\frac{23}{27}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

$V=\pi (x-\frac{x^3}{3(\frac{3}{2})^2})|_{-1}^{0}$

$V=\pi (0-\frac{0^3}{3(\frac{3}{2})^2})-\pi (-1-\frac{(-1)^3}{3(\frac{3}{2})^2})$

$V=\frac{23}{27}\pi$

Report an Error

Example Question #48 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of $\frac{1}{4}$ and a major axes (the horizontal length) of, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is $\frac{1}{10}$ units to the left of the origin and the second cut is $\frac{1}{10}$ units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{37}{3000}\pi$

$\frac{37}{150}\pi$

$\frac{37}{1500}\pi$

$\frac{37}{300}\pi$

$\frac{37}{750}\pi$

Correct answer:

$\frac{37}{3000}\pi$

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius $b=\frac{1}{4}$ . and the length of the spheroid 2a=1 , or $a=\frac{1}{2}$ :

$V=\pi (\frac{1}{4})^2(x-\frac{x^3}{3(\frac{1}{2})^2})|_{-\frac{1}{10}}^{\frac{1}{10}}$

$V=\pi (\frac{1}{4})^2((\frac{1}{10})-\frac{(\frac{1}{10})^3}{3(\frac{1}{2})^2})-\pi (\frac{1}{4})^2((-\frac{1}{10})-\frac{(-\frac{1}{10})^3}{3(\frac{1}{2})^2})$

$V=\frac{37}{3000}\pi$

Report an Error

Example Question #49 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 0.8 and a major axes (the horizontal length) of 2, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 0.6 units to the left of the origin and the second cut is 0.4 units to the left of the origin, what is the volume of the segment?

Possible Answers:

0.386

0.300

0.464

0.289

0.217

Correct answer:

0.300

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=0.8 . and the length of the spheroid 2a=2 , or a=1 :

$V=\pi (0.8)^2(x-\frac{x^3}{3})|_{-0.6}^{-0.4}$

$V=\pi (0.8)^2(-0.4-\frac{(-0.4)^3}{3})-\pi (0.8)^2(-0.6-\frac{(-0.6)^3}{3})$

V=0.300

Report an Error

Example Question #50 : Volume

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 0.3 and a major axes (the horizontal length) of 0.9 , has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 0.3 units to the left of the origin and the second cut is 0.1 units to the left of the origin, what is the volume of the segment?

Possible Answers:

0.067

0.033

0.078

0.044

0.056

Correct answer:

0.044

Explanation:

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=0.3 . and the length of the spheroid 2a=0.9 , or a=0.45 :

$V=\pi (0.3)^2(x-\frac{x^3}{3(0.45)^2})|_{-0.3}^{-0.1}$

$V=\pi (0.3)^2(-0.1-\frac{(-0.1)^3}{3(0.45)^2})-\pi (0.3)^2(-0.3-\frac{(-0.3)^3}{3(0.45)^2})$

V=0.044

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Calculus 1 : Regions

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