19–20. Area bounded by parametric curves Find the area of the ...

Question

19–20. Area bounded by parametric curves Find the area of the following regions. (Hint: See Exercises 103–105 in Section 12.1.) The region bounded by the y-axis and the parametric curve

The region bounded by the x-axis and the parametric curve x=cost, y=sin2t, for 0≤t≤π/2

Accepted Answer

Welcome back, everyone. Compute the area enclosed between the X axis and the parametric curve. X equals cosine C, Y equals sin of 2 T for T between 0 and 5 divided by 2 inclusive. Use the formula A equals integral from A to B of the absolute value of X T multiplied by Y T D T A 2/3 square units, B3 halves square units. C 8/3 square units, and D 2 square units. So for this problem, according to the formula, we're going to begin by identifying the limits of integration. The lower limit is A equals 0, and the upper limit B is 5 divided by 2. From here we want to find X T, which is the derivative of X with respect to T. So we want to find the derivative of cosine of T, which is negatives of T. And then we can use the formula, right? So we're going to show that the area is equal to the integral from 0 up to pi divided by 2 of the absolute value of negative sine of T multiplied by Y of T, which is of 2 T, right? DT And now we can essentially simplify. Let's remember the double angle formula for sine. We're going to get the integral from 0 up to pi divided by 2. Of the absolute value of negative sine C multiplied by. To sine C cosine T. So here we are applying the double angle identity, and then we can multiply sign by sign, right? So we get the integral from 0 up to pi divided by 2 of the absolute value of -2 squad of T, cosine of TDT. Now, notice that we can actually drop the absolute value sign because from 0 up to pi divided by 2 cosine is non-negative. And we also have sin squared, right? So, even though sine is non-negative as well, between 0 and pi divided by 2, we also have that square, and essentially, the absolute value of -2 is going to return 2. So, we can drop the absolute value. And the negative sign as well. We can also factor out 2, and we are going to integrate. We can integrate using the U substitution, and in particular, we can set a U equals sin T. Now, let's find the new limits of integration. The lower limit of integration U1 is going to be a sign of 0. Which is 0 and the upper limit of integration U2 is going to be sin. Of pi divided by 2. Which is one Now, let's convert DT into DU. So we can show that DU equals the derivative of sinus cosine. So we got the DU equals cosine of TDT. And therefore solving for the differential DC we can show that it is equal to. DU divided by cosine of T. So DU divided by cosine of T. And now we can go back into. The integral and we can show that area A equals 2 multiplied by the integral from 0 to 1 of U2, we have sin squared which becomes U2 and now cosine of T. Is multiplied by the T, which is the u divided by cosine of T. Notice that we can now cancel out the cosine, and we are then simply integrating U squared with respect to you. Let's go ahead and integrate. According to the power rule, we get 2, which is our constant multiplied by U cubed, divided by 3, and evaluated from 0 to 1. We got 2/3 in 1 cubed minus 0 cubed, which is equal to 2/3. So we got 2/3 square units, which corresponds to the answer choice A. We can label our final answer. Let's go ahead and do that and conclude the video. Thank you for watching.

19–20. Area bounded by parametric curves Find the area of the following regions. (Hint: See Exercises 103–105 in Section 12.1.) The region bounded by the y-axis and the parametric curve
The region bounded by the x-axis and the parametric curve x=cost, y=sin2t, for 0≤t≤π/2

Key Concepts

Parametric Equations and Curves

Area Under Parametric Curves

Limits of Integration for Parametric Curves

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