On or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Thus, if you are not sure content located Misrepresent that a product or activity is infringing your copyrights. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially Your Infringement Notice may be forwarded to the party that made the content available or to third parties such Means of the most recent email address, if any, provided by such party to Varsity Tutors. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by If Varsity Tutors takes action in response to Information described below to the designated agent listed below. Or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one If we were to use the Pythagorean Theorem, since we've already determined that b=h, that means a=b in the equation. Now that we have the base and the height, we can substitute the values into the area equation and get the triangle's area. Now to substitute in the value of c to solve for the height and base. That means the Pythagorean Theorem can be rewritten as: If we were to use the Pythagorean Theorem, since we've already determined that b=h, that mean a=b in the equation. The missing sides can be calulated in one of two ways: Therefore, we can logic out that the base and the height must be the same. This means that the hypotenuse value must be used to determine the height and the base.īecause this is a 45/45/90 triangle, this means that it is also isosceles. Given the area equation, the problem hasn't given any numbers that can be substituted into the equation to solve for an area. In the problem, the only information given is what type the triangle is and what its hypotenuse is. This entry was posted in Introductory Problems, Volumes by cross-section on Jby mh225.To find the area of a triangle, we must use where b=base and h=height. The cross-sections are circles of radius x 2, so the cross-sectional area is A(x) π⋅(x 2) 2π⋅x 4 The volume is V = ∫ -1 1A(x) dx = ∫ -1 1 π⋅x 4 dx = π⋅(x 5/5)| -1 1 = 2π/5 Find the volume of the solid obtained by rotating the curve y = x 2, -1 ≤ x ≤ 1, about the x-axis. where x⋅ex 2 was integrated using the substitution u = x 2, so du = 2xdx.ĥ. The area is A(x) = base ⋅ height = x⋅ex 2. Find the volume of the solid with cross-section a rectangle of base x and height e x 2 Answerġ. where cos(x)sin 2(x) is integrated using the substitution u = sin(x), so du = cos(x) dx.Ĥ. Find the volume of the solid with circular cross-section of radius cos 3/2(x), for 0 ≤ x ≤ π/2. Recall an ellipse with semi-major axis a and semi-minor axis b has area πab, so this ellipse with semi-major axis x 2 and semi-minor axis x 3 has the area: A(x) = π⋅x 2⋅x 3 = π⋅x 5. Find the volume if the solid with elliptical cross-section perpendicular to the x-axis, with semi-major axis x 2 and semi-minor axis x 3, for 0 ≤ x ≤ 1 Answerġ. Find the volume of the solid with right isosceles triangular cross-section perpendicular to the x-axis, with base x 2, for 0 ≤ x ≤ 1 Answerġ.
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