Calculus AB Homework 7.3 Area Between Curves

This paragraph introduces a video lesson on solving calculus problems to find areas between curves. The focus is on unit 7 homework problems 21 through 34, specifically using vertical and horizontal cross sections to find the area bounded by y=x^2 and y=x^3. The instructor explains the process of setting up integrals with respect to x (dx) and y (dy), and emphasizes that the area should be consistent regardless of the method used. The first example involves finding the area between the curves from x=0 to x=1 using a vertical cross section, calculating the area as 1/12. The process includes identifying the intersection points, setting up the integral, and evaluating it to find the area.

The second paragraph continues the discussion on finding the area between curves but introduces the horizontal cross section method. The instructor explains the process of converting the equations to be in terms of y to solve for x, which is necessary for integrating with respect to y (dy). The example provided involves the same curves as before but now with the integration limits and functions rearranged to reflect the horizontal approach. The instructor demonstrates the integration process and evaluates it to find the area, which matches the result from the vertical cross section method.

This paragraph delves into a more complex problem where the area between the curves y=√x and y=1/4x is to be found. The instructor first identifies the intersection points and then sets up the integral for a vertical cross section. The process involves integrating from x=0 to x=16, using the difference between the two functions as the height of the rectangles in the Riemann sum. The antiderivative is calculated, and the area is evaluated to be 32/3. The paragraph also attempts a horizontal cross section approach but acknowledges the complexity due to the need to solve for x in terms of y and the presence of multiple intersection points.

The fourth paragraph discusses the horizontal cross section method for a different set of curves, y=2-x^2 and y=x. The instructor identifies the intersection points and sets up two integrals due to the change in the right function at the point of intersection. The integrals are set up from y=-2 to y=1 and from y=1 to y=2, with the respective functions for the right and left sides of the rectangle. The instructor uses a calculator to find the area, which turns out to be 9/2, after a brief discussion on the complexity of integrating the square root function.

The fifth paragraph presents a scenario where the horizontal cross section method is not straightforward due to the functions touching the same curve at different points. The instructor discusses the need to split the integral into two parts and to solve the equations for x in terms of y. The example involves the curves y=√x and y=x-2, with the integrals set up from y=-2 to y=0 and from y=0 to y=2. The area calculation is shown to be 16/3, demonstrating the process of integrating with respect to y and evaluating the results.

In this paragraph, the instructor advises against using vertical cross sections for certain problems due to the complexity of having the same function on top and bottom, which would require solving for y. The example given involves the curves x=1/2y^2 and y=-1/2x+3, and the instructor recommends using horizontal cross sections instead. The intersection points are identified, and the functions are rewritten in terms of x for the horizontal cross section. The integral is set up from y=-6 to y=2, and the area is calculated to be 44 2/3.

The instructor discusses how to find intersection points for horizontal cross sections, using the curves y=-1/2x+3 and x=1/2y^2 as an example. The functions are labeled, and the intersection points are calculated to be (2,2) and (18,-6). The integral is set up from y=-6 to y=2, and the right and left functions are identified for the horizontal cross section. The area is calculated using the difference between the functions, resulting in an area of 32/3.

This paragraph involves identifying intersection points and calculating areas for the curves y=x^2 and y=-4x. The instructor finds the intersection points to be (-4,16) and (0,0) and sets up the integral for a vertical cross section from x=-4 to x=0. The integral involves the difference between the two functions, and the area is calculated to be 32/3. The horizontal cross section method is also discussed, with the integral set up from y=0 to y=16, and the area is found to be the same as with the vertical method.

The instructor tackles a complex problem involving the curves x=1/2y^2 and y=-1/2x+3, which require careful identification of intersection points and the correct setup of integrals for horizontal cross sections. The intersection points are found to be (2,2) and (18,-6). The integral is split into two parts due to the change in the function at the intersection point. The area calculation involves integrating the difference between the functions from y=-6 to y=2, resulting in an area of 42 2/3.

The final paragraph involves finding the value of K for which the area under the curve y=1/X from x=2 to x=K is equal to the natural logarithm of 4. The instructor sets up the integral for a vertical cross section from x=2 to x=K and evaluates it to find the area. Using properties of logarithms, the equation is simplified to find that the absolute value of K is 8, leading to the conclusion that K is 8, assuming K is greater than 2.

This paragraph discusses finding the value of C that divides the area under the curve y=cosine(X) between x=0 and x=π/2 into two equal parts. The instructor sets up two integrals, one from x=0 to x=C and another from x=C to x=π/2, and equates them to find C. By solving the resulting trigonometric equation, the value of C is found to be π/6, which corresponds to an angle of 30 degrees in Quadrant 1.

The instructor calculates the area enclosed by the y-axis, the curve y=3cosine(X), and the line y=X in the first quadrant. Using a calculator, the intersection point is found to be at (1.7, 3cosine(1.7)). The integral is set up as a DX problem from x=0 to x=1.7, integrating the difference between the two functions. The resulting area is calculated to be approximately 2.078, demonstrating the process of using a calculator to solve integrals for area calculations.