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Applications – Volumes of Revolution II
Problem 243: Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. (1) about the -axis. Solution: The volume of revolution is given by (2) where is the distance between the two curves. We are not given the limits of integration. We need…
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Applications – Volumes of Revolution
Problem 242: Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. (1) about the -axis. Solution: The volume of revolution is given by (2) where is the distance between the two curves. We are given one limit of integration . We need…
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Applications Surface Area II
Problem 241: Find the area of the surface generated by revolving the given curve about the -axis. (1) Solution: The surface area by revolving around the -axis is given by (2) where is the distance between the axis and the given curve, and and are the limits of integration. Let start by finding…
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Applications Surface Area
Problem 240: Find the area of the surface generated by revolving the given curve about the -axis. (1) Solution: The surface area by revolving around the -axis is given by (2) where is the distance between the axis and the given curve, and and are the limits of integration. Let start by finding…
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Limit of Special Trigonometric
Problem 239: Proof that (1) Solution: We will start by multiplying the numerator and denominator by the conjugate . That is, (2) Let’s recall that so that . Thus (3) This completes the proof.
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Integration by Parts (Reduction Formula)
Problem 238: Show that (1) Solution: One can use integration by parts to show this. That is, let (2) and (3) Using the integration by parts formula, (4) This completes the proof. (5)
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Integration by Parts (Reduction Formula)
Problem 237: Show that (1) Solution: One can use integration by parts to show this. That is, let (2) and (3) By the integration by parts formula, (4) This shows what we wanted to show. (5)
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Arc Length II
Problem 236: Find the arc length of the graph of the function from to . Solution: The arc length is given by (1) That is, (2) Thus, (3) Using U-substitution to evaluate the integral, we get (4) Therefore, the arc length of the graph between 2 and 3 is approximately …
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Arc Length
Problem 235: Find the length of the curve defined by (1) from to . Solution: The length of a curve is given by (2) That is, (3) Then (4) That is, the length of the curve between 0 and 1 is
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Integration – Partial Fractions
Problem 234: Evaluate the integral (1) Solution: First, we are going to use long division to rewrite our integral. That is, using long division, (2) One can now use this to rewrite our integral and integral. Thus, (3) That is, (4)
One Problem at a Time
